Topics

The acceleration due to gravity is given by

\[g = \frac{F}{\text{ }m} = \frac{GM}{R^{2}}\]

where \(F\) is the force exerted by the earth on an object of mass \(m\). This force is affected by a number of factors and hence g also depends on these factors.

(a) height from the surface of the Earth

If the object is placed at a distance \(h\) above the surface of the earth, the force of gravitation on it due to the earth is

\[F = \frac{GMm}{(R + h)^{2}}\]

where \(M\) is the mass of the earth and \(R\) is its radius.
Thus, \(\ g = \frac{F}{m} = \frac{GM}{(R + h)^{2}}\)
We see that the value of \(g\) decreases as one goes up. We can write,

\[g = \frac{GM}{R^{2}\left( 1 + \frac{h}{R} \right)^{2}} = \frac{g_{0}}{\left( 1 + \frac{h}{R} \right)^{2}}\]

where \(g_{0} = \frac{GM}{R^{2}}\) is the value of g at the surface of the earth. If \(h \ll R\),

\[g_{0} = g_{0}\left( 1 + \frac{h}{R} \right)^{- 2} \approx \left( 1 - \frac{2\text{ }h}{R} \right).\]

If one goes a distance \(h\) inside the earth such as in mines, the value of \(g\) again decreases. The force by the earth is, by equation

\[\begin{matrix} F & \ = \frac{GMm}{R^{3}}(R - h) \\ \text{~}\text{or,}\text{~}\ g & \ = \frac{F}{\text{ }m} = \frac{GM}{R^{2}}\left( \frac{R - h}{R} \right) \\ g & \ = g_{0}\left( 1 - \frac{h}{\text{ }g} \right) \end{matrix}\]

The value of \(g\) is maximum at the surface of the earth and decreases with the increase in height as well as with depth.

Variation with latitude (Due to rotation of earth)

In figure \(m\) is a mass placed on a weighing machine situated at a latitude of \(\phi\) the real forces acting on it are :
(i) the gravitational force mg towards the center of earth.
(ii) the normal reaction N of the weighing machine directed away from the center of earth,
(iii) The horizontal reaction H of the weighing machine directed along the tangent as shown.

In the reference frame fixed to the earth's surface the body would also be acted upon by the pseudo force (centrifugal force) \({mw}^{2}r\) directed as shown.
where \(r\) is the distance of \(P\) from earth's axis, \(r = R_{e}cos\phi\), where \(R_{e}\) is radius of earth.
Now in this reference frame \(m\) is at rest. Resolving forces along the radial and normal direction we get,

\[\begin{matrix} mg & \ = m\omega^{2}rcos\phi + N \\ & \ = m\omega^{2}R_{e}\cos^{2}\phi + N \Rightarrow \text{ }N = mg - m\omega^{2}R_{e}\cos^{2}\phi \\ H & \ = m\omega^{2}R_{e}cos\phi sin\phi \end{matrix}\]

Now the effective weight of the body is the net force experienced by the weighing machine which is equal and opposite to the force exerted by the weighing machine on the body.
\[{\therefore m{\overrightarrow{g}}_{\text{eff}\text{~}} = - (\overrightarrow{N} + \overrightarrow{H}) = - (N\widehat{n} + H\widehat{t}) }{\Rightarrow {\overrightarrow{g}}_{\text{eff}\text{~}} = - g\left\lbrack \left( \frac{1 - \omega^{2}R_{c}\cos^{2}\phi}{\text{ }g} \right)\widehat{n} - \frac{\omega^{2}R_{e}cos\phi sin\phi\widehat{t}}{g} \right\rbrack }\]Now \(\frac{\omega^{2}R_{e}}{g} \simeq 0.0337\), Hence its square may be neglected and
thus \(g_{\text{eff}\text{~}} \simeq g - \omega^{2}R\cos^{2}\phi\)
\(g_{\text{eff}\text{~}} =\) apparant according due to gravity at a latitude of \(\phi\)

• At poles, \(\phi = \pi/2\) and hence \(g_{\text{eff}\text{~}} = g\)

• At the equator, \(\phi = 0\) and hence \(g_{\text{eff}\text{~}} = g - \omega^{2}R\)

(c) Nonsphericity of the Earth

All formulae and equations have been derived by assuming that the earth is a uniform solid sphere. the shape of the earth slightly deviates from the perfect sphere. The radius in the equatorial plane is about 21 km larger than the radius along the poles. Due to this the force of gravity is more at the poles and less at the equator. The value of \(g\) is accordingly larger at the poles and less at the equator. Note that due to rotation of earth also, the value of \(g\) is smaller at the equator than that at the poles.

(d) Nonuniformity of the Earth

The earth is not a uniformly dense object. There are a veriety of minerals, metals, water, oil, etc., inside the earth. Then at the surface there are mountains, seas, etc. Due to these nonuniformities in the mass distribution, the value of \(g\) is locally affected.

The exerted force by the earth on a body is called the weight of the body. In this sence "weight of the earth" is a meaningless concept. However, the mass of the earth can be determined by noting the acceleration due to gravity near the surface of the earth. We have,

\[g = \frac{GM}{R^{2}}\]

or, \(\ M = {gR}^{2}/G\)
Putting \(g = 9.8\text{ }m{\text{ }s}^{- 2},R = 6400\text{ }km\)
and \(G = 6.67 \times 10^{- 11}\frac{\text{ }N - m^{2}}{{\text{ }kg}^{2}}\)
the mass of the earth comes out to be \(5.98 \times 10^{- 24}\text{ }kg\).

Practice Exercise

Q. 1 Earth's mass is 80 times that of the moon and their diameters are 12800 and 3200 kms respectively. What is the value of \(g\) at the moon? \(g\) on earth \(= 980\text{ }cm/s^{2}\).
Q. 2 The diameter of a planet is four times that of the earth. Find the time period of pendulum on the planet, if it is a second pendulum on the earth. Take the mean density of the planet equal to that of the earth,
Q. 3 A tunnel is dug along a chord of the earth at perpendicular distance \(R/2\) from the earth's centre. The wall of the tunnel may be assumed to be frictionless. Find the force exerted by the wall on particle of mass \(m\) when it is at a distance \(x\) from the centre of the tunnel.
Q. 4 Find the height over earth's surface at which the weight of a body becomes half of its value at the surface.

Answers

Q. \(1196\text{ }cm/s^{2}\)
Q. 21 s
Q. \(3\ \frac{{GM}_{e}m}{2R^{2}}\)
Q. \(4(\sqrt{2} - 1)\)

In analyzing the motion of planets and satellites, it is often easier and more informative to use energy rather than force. In this section we shall evaluate the potential energy of a system consisting of two bodies that interact through the gravitational force. we obtained the potential energy change due to gravity for a body that moves through a height \(y\) near the Earth's surface : \(\Delta U = mgy\). However, this applies only near the Earth's surface, where (for changes in height that are small compared with the distance from the center of the Earth) we can regard the gravitational force as approximately constant. Our goal here is to find a general expression that applies at all location, such as at the altitude of an orbiting satellite.

The potential energy difference can be found from equation. \(\Delta U = U_{b} - U_{a} = - W_{ab}\), where \(W_{ab}\) is the work done to configuration b. However, this equation applies only if the force is conservative.

Calculating the Potential Energy

The gravitational force is conservative so we can calculate the potential energy. Figure shows a particle of mass m moving from a to b along a radial path. A particle of mass M , which we assume to be at rest at the origin, exerts a gravitational force on m . The vector \(\overrightarrow{r}\) locates the position of m relative to M by the gravitational force is

\[\begin{matrix} & W_{ab} = - \int_{a}^{b}\mspace{2mu}\mspace{2mu} Fdr \\ & \ = - \int_{r_{a}}^{r_{b}}\mspace{2mu}\mspace{2mu}\frac{GMm}{r^{2}}dr = - GMm\int_{r_{3}}^{r_{b}}\mspace{2mu}\mspace{2mu}\frac{dr}{r^{2}} \\ & \ = - \left. \ GMm\left( - \frac{1}{r} \right) \right|_{r_{a}}^{r_{b}} = GMm\left( \frac{1}{r_{b}} - \frac{1}{r_{a}} \right) \end{matrix}\]

The negative sign in the first line of this equation arises because the (attractive) force \(\overrightarrow{F}\) and the infinitesimal radial vector \(d\overrightarrow{r}\) point in opposite directions. Equation shows that, when \(r_{b} > r_{a}\) (as in figure), the work \(W_{ab}\) is negative, as we expect.
Applying equation ( \(\Delta U = - W_{ab}\) ), we can find the change in the potential energy of the system as \(m\) oves between points a and b
\[\Delta U = U_{b} = - W_{ab} = GMm\left( \frac{1}{r_{a}} - \frac{1}{r_{b}} \right) \]A particle of mass \(M\) exerts a gravitational force \(\overrightarrow{F}\) on a particle of mass \(m\) that moves from a to \(b\).
If \(m\) moves outward from \(a\) to \(b\), the change in potential energy is positive \(\left( U_{b} > U_{a} \right)\). That is, if the particle passes through point a with a certain kinetic energy \(K_{a}\), as it travels to b its gravitational potential energy increases as its kinetic energy decreases \(\left( K_{b} < K_{a} \right)\). Conversely, if the particle is moving inward, its potential energy decreases as its kinetic energy increases.

Instead of differences in potential energy, we can consider the value of the potential energy at a single point if wedefine a reference point. We choose our reference configuration to be an infinite separation of the particles, and the we define the potential energy to be zero in that configuration. Let us evaluate equation for \(r_{b} = \infty\) and \(U_{b} = 0\). If a represents any arbitrary point, where the separation between the particles is \(r\), then equation becomes

\[U(\infty) - U(r) = GMm\left( \frac{1}{r} - 0 \right)\]

or

\[U(r) = - \frac{GMm}{r}\]

Note :

1. We can reverse the previous calculation and derive the gravitational force from the potential energy. For spherically symmetric potential energy functions, the relation \(F = - dU/dr\) gives the radial component of the force; see equation. With the potential energy of equation, we obtain
   \[F = - \frac{dU}{dr} = - \frac{d}{dr}\left( - \frac{GMm}{r} \right) = - \frac{GMm}{r^{2}} \]The minus sign here shows that the force is attractive, directed inward along a radius.

2. We can show that the potential energy defined according to equation leads to the familiar mgy for a small difference in elevation y near the surface of the Earth. Let us evluate equation for the difference in potential energy between the location at a height \(y\) above the surface (that is, \(r_{b} = R_{E} + y\), where \(R_{E}\) is the radius of the Earth) and the surface ( \(r_{a} = R_{E}\) )
   \[{\Delta U = U\left( R_{E} + y \right) - U\left( R_{E} \right) = {GM}_{E}m\left( \frac{1}{R_{E}} - \frac{1}{R_{E} + y} \right) }{= \frac{{GM}_{E}m}{R_{E}}\left( 1 - \frac{1}{1 + y/R_{E}} \right) }\]When \(y \ll R_{E}\), which would be the case for small displacements of bodies near the Earth's surface, we can use the binomial expansion to approximate the last term as \((1 + x)^{- 1} = 1 - x + \ldots. \approx 1 - x\), which gives
   \[\Delta U \approx \frac{{GM}_{E}m}{R_{E}}\left\lbrack 1 - \left( 1 - \frac{y}{R_{E}} \right) \right\rbrack = \frac{{GM}_{E}my}{R_{E}^{2}} = mgy \]using equation to replace \({GM}_{E}/R_{E}^{2}\) with g .

3. If our system contains more than two particles, we consider each pair of particles in turn, calculate the gravitational potential energy of that pair with Equation as if the other particles were not there, and then algebraically sum the results. Each of the three pairs of Figure, for example, gives the potential energy of the system as

If our system contains more than two particles, we consider each pair of particles in turn, calculate the gravitational potential energy of that pair with Eq. below as if the other particles were not there, and then algebraically sum the results. the potential energy of the system as

\[U = - \left( \frac{Gm_{1}m_{2}}{r_{12}} + \frac{Gm_{1}m_{3}}{r_{13}} + \frac{Gm_{2}m_{3}}{r_{23}} \right)\]

Gravitational Potential

The gravitational potential at a point in gravitational field is the gravitational potential energy per unit mass placed at the point in gravitational field. Thus at a certain point in gravitational field, a mass \(m_{0}\) has a potential energy \(U\) the gravitational potential at that point is given as

\[V = \frac{U}{{\text{ }m}_{0}}\]

or if at a point in gravitational field gravitational potential V is known then the interaction potential energy of a point mass \(m_{0}\) at the point in the field is given as

\[U = m_{0}\text{ }V\]

Interaction energy of a point mass \(m_{0}\) in a field is defined as work done in bringing that mass from infinity to the point. In the same fashion we can define gravitational potential at a point in field, alternatively as "Work done in bringing a unit mass from infinity to that point against gravitational force."
When a unit mass is brought to a point in a gravitational field, force on the unit mass is \(\overrightarrow{g}\) at a point in the field. Thus the work done in bringing this unit mass from infinity to a point P in gravitational field or gravitational potential at point P is given as

\[V_{P} = - \int_{\infty}^{P}\mspace{2mu}\overrightarrow{\text{ }g} \cdot dx\]

Here negative sing shown that \(V_{P}\) is the negative of work done by gravitation field or it is the external required work for the purpose against gravitational force.

Gravitational Potential due to a Point Mass

We place a test mass \(m_{0}\) at \(P\) and we find the interaction energy of \(m_{0}\) with the field of \(m\), which is given as

\[U = - \frac{{Gmm}_{0}}{x}\]

Now the gravitational potential at P due to m can be written as

\[V = \frac{U}{{\text{ }m}_{0}} = - \frac{Gm}{r}\]

Gravitational potential due to a ring

At the centre of ring
In gravitational potential the situation is not like this as it is a scaler quantity and here the distane of centre from each element dm on ring circumference is equal to R , thus every element dm produces an equal gravitational potential at C , given as

\[dV = - \frac{Gdm}{R}\]

Now due to the whole ring the gravitational potential at its centre C is given as

\[V_{C} = \int_{}^{}\ dV = - \int_{}^{}\ \frac{Gdm}{R} = - \frac{Gm}{R}\]

Note : Now we have seen that most of the expressions of \(\overrightarrow{g}\) and \(V\) are same as that we have calculated in the topic electrostatics. So we will not prove them again rather we can replace some sysbols as :

For Formula conversion :

Body	Gravitational Field	Gravitational potential
Point mass	\[- \frac{Gm}{r^{2}}\]	\[- \frac{Gm}{r}\]
Ring	\[\frac{Gmx}{\left( R^{2} + x^{2} \right)^{3/2}};\left. \ \ E_{\text{max}\text{~}} \right|_{x = R/\sqrt{2}}\]	\[- \frac{Gm}{\sqrt{x^{2} + R^{2}}}\]
Disc	\[2\pi G\sigma\left( 1 - \frac{x}{\sqrt{x^{2} + R^{2}}} \right)\]	\[- 2\pi G\sigma\left\lbrack \sqrt{x^{2} + R^{2}} - x \right\rbrack\]
infinite wire	\[E = 2G\lambda/r\]	\[\Delta V = 2G\lambda ln\frac{r_{1}}{r_{2}}\]
wire	\[\frac{G\lambda L}{x(x + L)}\]	\[\lambda\left\lbrack \ ^{- G\lambda ln\left( \frac{x + L}{x} \right)} \right.\ \]
circular arc & rod	\[- \frac{G\lambda}{r}\begin{bmatrix} \left( sin\theta_{1} + sin\theta_{2} \right)\widehat{i} \\ - \left( cos\theta_{1} - cos\theta_{2} \right)\widehat{j} \end{bmatrix}\]
Apex of cone	\[\infty\]	\[- 2\pi G\sigma R\]
infinite plate	\[E = 2\pi G\sigma\]	\[\Delta V = 2\pi G\sigma d\]
Hollow Sphere	\[\overrightarrow{E} = \left\{ \begin{matrix} 0 & 0 \leq r < R \\ - \frac{GM}{r^{2}}\widehat{r} & r \geq R \end{matrix} \right.\ \]	\[V = \left\{ \begin{matrix} - \frac{GM}{R} & 0 \leq r < R \\ - \frac{GM}{r} & r \geq R \end{matrix} \right.\ \]
Solid sphere	\[\overrightarrow{E} = \left\{ \begin{matrix} - \frac{GM}{R^{3}} = - \frac{4\pi G\rho}{3}\overrightarrow{r} & 0 \leq r < R \\ - \frac{Kq}{r^{2}}\widehat{r} & r \geq R \end{matrix} \right.\ \]	\[V = \left\{ \begin{matrix} - \frac{GM}{2R}\left( 3 - \frac{r^{2}}{R^{2}} \right) & 0 \leq r < R \\ - \frac{GM}{r} & r \geq R \end{matrix} \right.\ \]

Practice Exercise

Q. 1 Find the kinetic energy needed to project a body of mass m from the centre of a ring of mass M and radius R so that it will never come back.
Q. 2 How much work is done in circulating a small object of mass \(m\) around a sphere of mass \(m\) in a circle of radius R.
Q. 3 Find the gravitational potential due to a hemispherical cup of mass M and radius R , at its centre of curvature.
Q. 4 Find the gravitational potential energy of system consisting of uniform rod AB of mass M , length \(l\) and a point mass \(m\) as shown in figure.

Answers

Q. \(1\ \frac{GMm}{R}\)
Q. \(2\ 0\)
Q. \(3 - \frac{\text{~}\text{GM}\text{~}}{\text{~}\text{R}\text{~}}\)
Q. \(4 - \frac{\text{~}\text{GMm}\text{~}}{l}In\left( 1 + \frac{l}{r} \right)\)

(a) Planets

Planets move round the sun due to the gravitational attraction of the sun. The path of these planets are elliptical with the sun at a focus.
(b) Satellite

Satellites are launched from the earth so as to move round it. A number of rockets are fired from the satellite at proper time to establish the satellite in the desired orbit. Once the satellite is placed in the desired orbit with the correct speed for that orbit, it will continue to move in that orbit under gravitational attraction of the earth.

We make two assumptions that simplify the analysis:
(1) We consider the gravitational force only between the orbiting body (the Earthy, for instance and the central body(the Sun), ignoring the perturbing effect of the gravitational force of other bodies (such as other planets)
(2) We assume that the central body is so much more massive than the orbiting body that we can ignore its motion under their mutual interaction.

Let the speed of an artificial earth satellite in its orbits of radius \(r\) be \(v_{0}\). The satellite is accelerating towards the centre of earth by earth's gravitational pull \(\frac{GMm}{r^{2}}\)

\[\begin{matrix} \Rightarrow & F_{CP} = \frac{GMm}{r^{2}} \\ & \frac{{mv}_{0}^{2}}{r} = \frac{GMm}{r^{2}} \\ \Rightarrow & v_{0} = \sqrt{\frac{GM}{r}} \end{matrix}\]

Putting \(\frac{GM}{r^{2}} = g\) (acceleration due to gravity at the orbit), we obtain,

\[\begin{array}{r} \Rightarrow \ v_{0} = \sqrt{gr}\#(ii) \end{array}\]

When it orbits at an altitude h , putting \(r = (R + h)\)
\[\Rightarrow \ v_{0} = \sqrt{\frac{GM}{R + h}} = \sqrt{\frac{GM}{R(1 + h/R)}} = \sqrt{\frac{gR}{(1 + h/R)}}\]

Angular speed

The angluar speed

\[\omega = \frac{V_{0}}{r}\]

Putting \(v_{0} = \sqrt{\frac{GM}{r}}\), we obtain

\[\omega = \sqrt{\frac{GM}{r^{3}}} = \sqrt{\frac{GM}{(R + h)^{3}}}\]

Angular momentum

The angular momentum of an earth satellite or a planet is given as
\[L = mvr\]

\[\begin{matrix} & \ = m\sqrt{\frac{GM}{r}} \times r \\ & \ = m\sqrt{GMr} \end{matrix}\]

Time period of Revolution

The period of revolution

\[T = \frac{2\pi}{\omega}\]

Puting \(\omega = \sqrt{\frac{GM}{r^{3}}}\)

\[T = 2\pi\sqrt{\frac{r^{3}}{GM}}\]

Energy consideration in planetary and satellite motion

\[U(r) = - \frac{GMm}{r}\]

Where \(r\) is the radius of the circular orbit.
The kinetic energy of the system is

\[K = \frac{1}{2}{mv}^{2} = \frac{1}{2}{\text{ }m}^{2}r^{2}\]

the Sun being at rest

\[\omega^{2}r^{2} = \frac{GM}{r}\]

so that (with \(v = \omega\) r)

\[K = \frac{GMm}{2r}\]

The total mechanical energy is

\[E = K + U = \frac{GMm}{2r} - \frac{GMm}{r} = \frac{GMm}{2r}\]

This energy is constant and negative. The kinetic energy can never be negative, but from equation we see that it

Note :

(1) Graph of kinetic energy K , potential energy U , and total energy \(E = K + U\) of a body in circular planetary motion.

A planet with total energy \(E_{0} < 0\) will remain in a orbit. The greater the distance from the Sun, the greater (that is, less negative) its total energy E .
(2) Gravitational binding energy

We have seen that if a particle of mass \(m\) placed on the earth is given an energy \(\frac{1}{2}m^{2} = \frac{GMm}{R}\) or more, it finally escapes from earth. The minimum energy needed to take the particle infinitely away from the earth is called the binding energy of the earth-particle system. Thus, the binding energy of the earthparticle system is \(\frac{GMm}{R}\)

Geostationary statellite :

A satellite which appears to be stationary when seen from earth is called a Geostationary satellite. For a satellite to be geostationary.
(i) Its orbit must be circular
(ii) It must rotate about the same axis as earth, i.e. it must move in the equatorial* plane.
(iii) It must revolve form west to east.
(iv) Its time period must be 24 hours.

\[\begin{matrix} \Rightarrow \ m\omega^{2}(R + h) & \ = \frac{GMm}{(R + h)^{2}} \\ \left( \frac{2\pi}{T} \right)^{2} & \ = \frac{GM}{(R + h)^{3}} \\ (R + h) & \ = \left\{ GM\left( \frac{T}{2\pi} \right)^{2} \right\}^{1/3}\ \text{~}\text{where}\text{~}T = 24 \times 3600sec. \end{matrix}\]

Solving \(h = 36000\text{ }km\) (approx)

• Any satellite must rotate about the center of earth.

[Orbital speed]

Where \(M\) is the mass of sun and \(r\) is the orbit radius of earth.
We know time period of earth around sun is \(T = 365\) days, thus we have

\[T = \frac{2\pi r}{v}\]

or

\[T = 2\pi r\sqrt{\frac{r^{3}}{GM}}\]

or

\[\begin{matrix} & M = \frac{4\pi^{2}r^{3}}{GT^{2}} \\ & \ = \frac{4 \times (3.14)^{2} \times (1.49 \times 1011)^{3}}{(365 \times 24 \times 3600)^{2} \times \left( 6.66 \times 10^{- 11} \right)} = 1.972 \times 10^{22}\text{ }kg \end{matrix}\]

Definition: If a particle of mass \(m\), kept in an attractive gravitational field is given sufficient kinetic energy, it may escape the gravitational pull due to the field. The particle will escape to infinity depending on whether its path allows it to do so. For example, a particle of mass \(m\) kept on the surface of the earth requires a minimum velocity \(v_{esc}\) so that it moves to infinity.

\[\frac{1}{2}{mv}_{esc}^{2} - \frac{GMm}{R} = 0\]

Where \(M\) is the mass of the earth and \(R\) is its radius.

\[v_{esc} = \sqrt{\frac{2GM}{r}}\]

Note :

The escape speed does not depend on the direction in which the projectile is fired. The Earth's rotation which we have not considered in this calculation does play a role, however. Firing eastward has an advantage in that the Earth's tangential surface speed, which is \(0.46\text{ }km/s\) at Cape Canaveral, provides part of the kinetic energy needed for escape, and thus less thrust from the rocket engines would be required to escape the Earth's gravity.

Practice Exercise

Q. 1 Two satellites \(A\) and \(B\) of the same mass are orbiting the earth at altitudes \(R\) and \(3R\) respectively, where \(R\) is the radius of the earth. Taking their orbits to be circular, obtain the ratios of their kinetic and potential energies.
Q. 2 A satellite is to revolve round the earth in a circle of radius 8000 km . With what speed should this satellite be projected into orbit? What will be the time period? Take \(g\) at the surface \(= 9.8\text{ }m./s^{2}\) and radius of the earth \(= 6400\text{ }km\).
Q. 3 A satellite of mass \(2 \times 10^{3}\text{ }kg\) has to be shifted from an orbit of radius 2 R to another of radius 3 R , where \(R\) is the radius of the earth. Calculate the minimum energy required.

Answers

Q. 12 : 1
Q. \(2\ 7.08\text{ }km/s.118\) minutes
Q. \(31.04 \times 10^{10}\text{ }J\)

The empirical basis for understanding the motions of the planets is there laws deduced by Kepler ( 1571 - 1630, well before Newton) from studies of the motion of the planet Mars.

1. The law of orbits: All planets move in elliptical orbits having the Sun at one focus. Newton was the first realize that there is a direct mathematical relationship between inverse-square ( \(1/r^{2}\) ) force and elliptical orbits. Figure shows a typical elliptical orbit.

A planet of mass \(m\) moving in a elliptical orbit around the Sun. The Sun, of mass \(M\), is at one focus of the elipse. F' marks the other or "empty" focus. The semimajor axis a of the ellipse, the perihelion distance \(R_{p}\), and the aphelion distance \(R_{e}\) are also shown. The distance \(ae\) locates the focal points, e being the eccentricity of the orbit.
other planets in the solar system, the eccentricities are small and the orbits are nearly circular.
The maximum distance \(R_{a}\) of the orbiting body from the central body is indicated by the prefix apo - (or sometimes ap-), as in aphelion (the maximum distance from the Sun) or apogee ( the maximum distance from Earth). Similarly, the cloest distance \(R_{p}\) is indicated by the prefix peri-, as in perihelion or perigee. As you can see from figure \(R_{a} = a(1 + e)\) and \(R_{p} = a(1 - e)\). For circular orbits. \(R_{a} = R_{p} = a\)
2. The Law of Areas: A line joining any planet to the Sun sweeps out equal area in equal times. Figure illustrates this law : in effect it says that the orbiting body moves more rapidly when it is close to the central body than it does when it is far aways. We now show that the law of areas is identical with the law of conservation of angular momentum.

Consider the small area increament \(\Delta A\) covered in a time interval \(\Delta t\), as shown in figure. The area of this approximately triangular wedge in one-half its base, \(r\Delta\theta\), times its height r . The rate at which this area is swept out is \(\Delta A/\Delta t = \frac{1}{2}(r\Delta\theta)(r)/\Delta t\). In the instantaneous limit this becomes

\[\frac{dA}{dt} = {Lim}_{\Delta \rightarrow 0}\frac{\Delta\text{ }A}{\Delta t} = {Lim}_{\Delta \rightarrow 0}\frac{1}{2}r^{2}\frac{\Delta\theta}{\Delta t} = \frac{1}{2}r^{2}\omega\]

Assuming we can regard the more massive body M as at rest, the angular momentum of the orbiting body m relative

(a) The equal shaded areas are covered in equal times by a line connecting the planet to the Sun, demonstrating the law of areas.
(b) The area \(\Delta A\) is covered in a time \(\Delta t\), during which the line sweeps through an angle \(\Delta\theta\).
to the origin at the central body is, according to equation \(L_{z} = I\omega = {mr}^{2}\omega\) (choosing the z axis perpendicular to the plane of the orbit). Thus

\[\frac{dA}{dt} = \frac{L_{z}}{2\text{ }m}\]

If the system of M and m is isolated, meaing that there is no net external torque on the system, the \(L_{z}\) is a constant; therefore \(dA/dt\) is also constant. That is in every interval dt in the orbit, the line connecting m and M sweeps out equal areas dA, which verifies kepler's second law. The speeding up of a comet as it passes close to the Sun is an example of this effect and is thus a direct consequence of the law of conservation of angular momentum.
3. The law of Periods: The square of the period of any planet about the Sun is proportional to the cube of seminajor axis of the. Let us prove this result for circular orbits. The gravitational force provides the necessary centripetal acceleration for circular motion.

If ' T ' is the period of revolution and ' a ' be the semi-major axis of the path of planet then according to kepler's III Law, we have

\[T^{2} \propto a^{3}\]

For circular orbits, it is a special case of ellipse when its major and minor axis are equal. If a planet is in a circular orbit of radius \(R\) around the sun then its revolution speed must be given as

\[v = \sqrt{\frac{GM_{s}}{r}}\]

Where \(M_{s}\) is the mass of sun. There you can recall that this speed is independent from the mass of planet. Here the time period of revolution can be given as

\[T = \frac{2\pi r}{v}\]

or

\[\begin{array}{r} T = \frac{2\pi r}{\sqrt{\frac{{GM}_{s}}{r}}}\#(A) \end{array}\]

Squaring equation no. (A) we get

\[\begin{array}{r} T^{2} = \frac{4\pi^{2}}{{GM}_{S}}r^{3}\#(B) \end{array}\]

Equation (B) verifies the statement of Kepler's third law for circular orbits. Similarly we can also verify it from elliptical orbits. For this we start from the relation we've derived earlier for rate sweeping area by the position vector of planet with respect to sun which is given as

\[\frac{dA}{dt} = \frac{L}{2\text{ }m}\]

Where L is the total angular momentum of planet during its motion consider the path of planet shown in figure is an elliptical path with sun on focus (-ae, 0 ).

Here \(r_{1}\) and \(r_{2}\) are the shortest and farthest distance of planet from sun during its motion, which are given as

\[\begin{matrix} & r_{1} = a(1 - e) \\ & r_{2} = a(1 + e) \end{matrix}\]

and
Where \(e\) is the centricity. From geometry we know that the relation in semi major axis and semiminor axis be is given as

\[b = a\sqrt{1 - e^{2}}\]

If \(v_{1}\) and \(v_{2}\) are the planet speeds at perihelion and aphelion points then from conservation of momentum we have

\[L = {mv}_{1}r_{1} = {mv}_{2}r_{2}\]

From energy conservation we have
or

\[\begin{matrix} & \frac{1}{2}m_{1}v_{1}^{2} - \frac{GM_{S}m}{r_{1}} = \frac{1}{2}mv_{2}^{2} - \frac{GM_{s}m}{r_{2}} \\ & v_{1}^{2} - v_{2}^{2} = 2GM_{S}\left\lbrack \frac{1}{r_{1}} - \frac{1}{r_{2}} \right\rbrack \\ & v_{1}^{2}\left\lbrack 1 - \frac{r_{1}^{2}}{r_{2}^{2}} \right\rbrack = 2GM_{S}\left\lbrack \frac{r_{2} - r_{1}}{r_{1}r_{2}} \right\rbrack \end{matrix}\]

or

\[v_{1} = \sqrt{\frac{2GM_{S}r_{2}}{\left( r_{1} + r_{2} \right)r_{1}}}\]

From equation (vi) and (vii) we have

\[v_{1} = \sqrt{\frac{{GM}_{S}}{a}\left( \frac{1 + e}{1 - e} \right)}\]

Now from equation(v) we have the total area of ellipse traced by the planet is given as
or

\[A = \frac{L}{2\text{ }m}\text{ }T\]

or

\[T = \frac{2\text{ }m}{\text{ }L}\text{ }A = \frac{2\text{ }m\pi ab}{\text{ }L} = \frac{2{\text{ }m}^{2}ab}{{mv}_{1}r_{l}}\]

or

\[T = \frac{2\text{ }m\pi a\left\lfloor a\sqrt{1 - e^{2}} \right\rfloor}{m\left\lbrack \sqrt{\frac{{GM}_{S}}{a}\left( \frac{1 + e}{1 - e} \right)\lbrack a(1 - e)\rbrack} \right\rbrack}\]

or

\[T^{2} = \frac{4\pi^{2}}{{GM}_{s}}a^{3}\]

For a body being projected tangentially from above earth's surface, say at a distance \(r\) from earth's center, the trajectory would depend on the velocity of projection v.

Velocity

1. velocity, \(v < \sqrt{\frac{GM}{r}\left( \frac{2R}{r + R} \right)}\)

2. \(\sqrt{\frac{GM}{r}} > v > \sqrt{\frac{GM}{r}\left( \frac{2R}{r + R} \right)}\)

3. Velocity is equal to the critical velocity of the orbit, \(v = \sqrt{\frac{{GM}_{e}}{r}}\)

4. Velocity is between the critical and escape velocity of the orbit

\[\sqrt{\frac{2{GM}_{e}}{r}} > v > \sqrt{\frac{{GM}_{e}}{r}}\]

5. \(v = v_{esc} = \sqrt{\frac{2{GM}_{e}}{r}}\)

6. \(v > v_{esc} = \sqrt{\frac{2{GM}_{e}}{r}}\)

Orbit

Body returns to earth following ellipitical Path.

Body acquires an elliptical orbit with earth as the far-focus w.r.t. the point of projection.

Circular orbit with radius r

Body acquires an elliptical orbit with earth as the near focus w.r.t. the point of projection.

Body just escapes earth's gravity, along a parabolic path.

Body escape earth's gravity along a hyperbolic path.

Practice Exercise

Q. 1 A planet of mass \(M\) moves around the sun along an ellipse so that its minimum distance from the sun is equal to \(r\) and the maximum distance to R. Making use of Kepler's law, find the its period of revolution around the sun.
Q. 2 What should be the orbit radius of a communication satellite so that it can cover \(75\%\) of the surface area of earth during its revolution.

Practice Exercise

Q. \(1\sqrt{\frac{(r + R)^{3}}{2{GM}_{s}}}\)
Q. \(21.15R_{e}\)

Now as we know the height of a geostationary satellite we can easily find the area of earth exposed on the satellite or area of the region in which the communication can be mode using this satelline.
Figure shows earth and its exposed area to a geostationary satellite. Here the angle \(\theta\) can be given as

\[\theta = \cos^{- 1}\left( \frac{R_{e}}{R_{e} + h} \right)\]

Now we can find the solid angle \(\Omega\) which the exposed area subtend on earth's centre as

\[\begin{matrix} \Omega & \ = 2\pi(1 - cos\theta) \\ & \ = 2\pi\left( 1 - \frac{R_{e}}{R_{e} + h} \right) = \frac{2\pi\text{ }h}{R_{e} + h} \end{matrix}\]

Thus the area of earth's surface to geostationary satellite is

\[S = \Omega R_{e}^{2} = \frac{2\pi\text{ }hR_{e}^{2}}{R_{e} + h}\]

Q. 1 A satellite is revoling round the earth in a circular orbit a rasius a with velocity \(v_{0}\). A particle is projected from the satellite in forward direction with relative velocity \(v = (\sqrt{5/4} - 1)v_{0}\). Calculate, during subsequent motion of the particle its minimum and maximum distance from earth's centre.
Sol. The corresponding situation is shown in figure

Initial velocity of statellite \(\ v_{0} = \sqrt{\left( \frac{GM}{a} \right)}\)
When particle is thrown with the velocity v relative to satellite, the resultant velocity of particle will become

\[\begin{matrix} & v_{R} = v_{0} + v \\ & \ = \sqrt{\left( \frac{5}{4} \right)}v_{0} = \sqrt{\left( \frac{5}{4}\frac{GM}{a} \right)} \end{matrix}\]

As the particle velocity is greater than the velocity requried for ciruclar orbit, hence the particle path deviates from circular path to elliptical path. At positions of minimum and maximum distance velocity vector are perpendicular to instantaneous radius vector. In this elliptical path the minimum distance of particle from earth's centre is a and maximum speed in the path is \(v_{R}\) and let the maximum distance and minimum speed in the path is \(r\) and \(v_{1}\) respectively.
Now as angular momentum and total energy remain conserved. Applying the law of conservation of angular momentum, we have
or

\[\begin{matrix} {mv}_{1}r = m\left( v_{0} + v \right)a & \lbrack\text{ }m = \text{~}\text{mass of particle}\text{~}\rbrack \\ v_{1} = \frac{\left( v_{0} + v \right)a}{r} & \\ = \frac{a}{r}\left\lbrack \sqrt{\left( \frac{5}{4}\frac{GM}{a} \right)} \right\rbrack & \\ = \frac{1}{r}\left\lbrack \sqrt{\left( \frac{5}{4} \times GMa \right)} \right\rbrack & \end{matrix}\]

Applying the law of conservation of energy

\[\frac{1}{2}mv_{1}^{2} - \frac{GMm}{r} = \frac{1}{2}m\left( v_{0} + v \right)^{2} - \frac{GMm}{a}\]

or

\[\begin{matrix} & \frac{1}{2}\text{ }m\left( \frac{5}{4}\frac{GMa}{r^{2}} \right) - \frac{GMm}{r} = \frac{1}{2}\text{ }m\left( \frac{5}{4}\frac{GM}{a} \right) - \frac{GMm}{a} \\ & \frac{5}{8} \times \frac{a}{r^{2}} - \frac{1}{r} = \frac{5}{8} \times \frac{1}{a} - \frac{1}{a} = \frac{3}{8a} \\ & 3r^{2} - 8ar + 5a^{2} = 0 \end{matrix}\]

or
or

\[r = a\text{~}\text{or}\text{~}\frac{5a}{3}\]

Thus minimum disatnce of the particle \(= a\)
And maximum distance of the particle \(= \frac{5a}{3}\)
Q. 2 A satellite is revolving around a planet of mass M in an elliptic orbit of semimajor axis a. Show that the orbital speed the satellite when it is at a distance r from the focus will be given by :

\[v^{2} = GM\left\lbrack \frac{2}{r} - \frac{1}{a} \right\rbrack\]

Sol. As in case of elliptic orbit with semi major axis a, of a satellite total mechanical energy remains constant, at any position of satellite in the orbit, given as

\[E = - \frac{GMm}{2a}\]

or

\[\begin{array}{r} KE + PE = - \frac{GMm}{2a}\#(i) \end{array}\]

Now, if at postion \(r,v\) is the orbital speed of satellite, we have

\[\begin{array}{r} KE = \frac{1}{2}{mv}^{2}\text{~}\text{and}\text{~}PE = - \frac{GMm}{r}\#(ii) \end{array}\]

so from equation (i) and (ii), we have

\[\frac{1}{2}mv^{2} - \frac{GMm}{r} = - \frac{GMm}{2a}\text{, i.e.}\text{~}\ v^{2} = GM\left\lbrack \frac{2}{r} - \frac{1}{a} \right\rbrack\]

Q. 3 A planet of mass \(m\) moves along an ellipse around the sun so that its maximum and minimum distance from the sun are equal to \(r_{1}\) and \(r_{2}\) respectively. Find the angular momentum of this plane relative to the centre of the sun.
Sol. If \(v_{1}\) and \(v_{2}\) are the velocity or planet at its apogee and perigee respectively then according to conservation of angular momentum, we have
or

\[\begin{matrix} & mv_{1}r_{1} = mv_{2}r_{2} \\ & v_{1}r_{1} = v_{2}r_{2} \end{matrix}\]

As the total energy of the planet is also constant, we have

\[- \frac{GMm}{r_{1}} + \frac{1}{2}mv_{1}^{2} = - \frac{GMm}{r_{2}} + \frac{1}{2}mv_{2}^{2}\]

Where \(M\) is the mass of the sun.
or

\[GM\left\lbrack \frac{1}{r_{1}} - \frac{1}{r_{2}} \right\rbrack = \frac{v_{2}^{2}}{2} - \frac{v_{1}^{2}}{2}\]

or

\[GM\left( \frac{r_{1} - r_{2}}{r_{1}r_{2}} \right) = \frac{v_{1}^{2}r_{1}}{2r_{2}^{2}} - \frac{v_{1}^{2}}{2}\]

or

\[\begin{matrix} GM\left( \frac{r_{1} - r_{2}}{r_{1}r_{2}} \right) & \ = \frac{v_{1}^{2}}{2}\left( \frac{r_{1}^{2}}{r_{2}^{2}} - 1 \right) \\ & \ = \frac{v_{1}^{2}}{2}\left( \frac{v_{1}^{2} - v_{2}^{2}}{v_{2}^{2}} \right) \end{matrix}\]

or

\[= v_{1}^{2} = \frac{2GM\left( r_{1} - r_{2} \right)r_{2}^{2}}{r_{1}r_{2}\left( r_{1}^{2} - r_{2}^{2} \right)} = \frac{2Gr_{2}}{r_{1}\left( r_{1} + r_{2} \right)}\]

or

\[v_{1} = \sqrt{\left\lbrack \frac{2GMr_{2}}{r_{1}\left( r_{1} + r_{2} \right)} \right\rbrack}\]

Now Angular momentum of planet is given as

\[\begin{matrix} & L = {mv}_{1}r_{1} \\ & \ = m\sqrt{\left\lbrack \frac{2{GMr}_{1}r_{2}}{\left( r_{1} + r_{2} \right)} \right\rbrack} \end{matrix}\]

Q. 4 A spaceship is sent to investigate a planet of mass M and radius R . While hanging motionless in space at a distance 5 R from the center of the planet, the spaceship fires an instrument package with speed \(v_{0}\) as shown in the figure. The packages has mass \(m\), which is much smaller than the mass of the spaceship. For what angle \(\theta\) will the package just graze the surface of the planet?

Sol. Let the speed of the instrument package is v when it grazes the surface of the planet.

Conserving angular momentum of the package about the centre of the planet \({mv}_{0} \times 5Rsin(\pi - \theta) = mvRsin90^{\circ}\)

Conserving mechanical energy
\[{- \frac{GMm}{5R} + \frac{1}{2}{mv}_{0}^{2} = - \frac{GMm}{R} + \frac{1}{2}{mv}^{2} }{\Rightarrow \frac{1}{2}\text{ }m\left( v^{2} - v_{0}^{2} \right) = \frac{4GMm}{5R} }{v^{2} - v_{0}^{2} = \frac{8GM}{5R} }\]substituting the value of \(v\) from equation (I) in equation (ii)
\[25v_{0}^{2}\sin^{2}\theta - v_{0}^{2} = \frac{8GM}{5R} \Rightarrow sin\theta\frac{1}{5}\sqrt{1 + \frac{8GM}{5v_{0}^{2}R}} \]or \(\theta = \sin^{- 1}\left\lbrack \frac{1}{5}\sqrt{\left( 1 + \frac{8GM}{5v_{0}^{2}R} \right)} \right\rbrack\)
Q. 5 A missile is launched at an angle of \(60^{\circ}\) to the vertical with a velocity \(\sqrt{0.75gR}\) from the surface of the earth ( R is the radius of the earth). Find its maximum height from the surface of earth. (Neglect air resistance and rotation of earth.)
From conservation of mechanical energy

\[\begin{array}{r} \frac{1}{2}{mv}_{0}^{2} - \frac{GMm}{R} = \frac{1}{2}{mv}_{1}^{2} - \frac{GMm}{R + h}\#(i) \end{array}\]

Also from conservation of angular momentum

\[\begin{array}{r} {mv}_{0}Rsin60^{\circ} = {mv}_{1}(R + h)\#(ii) \end{array}\]

Solving (i) and (ii) and putting \(v_{0} = \sqrt{\frac{3GM}{4R}}\), we get \(h \approx 0.25R\).

Q. 6 Find the minimum colatitude which can directly receive a signed from a geostationary satellite.

Sol. The farthest point on earth, which can receive signals from the parking orbit is the point where a length is drawn on earth surface from satellite as shown in figure

The colatitude \(\lambda\) of point P can be obtained from figure as

\[sin\lambda = \frac{R_{e}}{R_{e} + h} \simeq \frac{1}{7}\]

We know for a partking orbit \(h \simeq 6R_{e}\)

Thus we have

\[1 = \sin^{- 1}\left( \frac{1}{7} \right)\]

Q. 7 A satellite of mass \(m\) is orbiting the earth in a circular orbit of radius \(r\). It starts losing energy slowly at a constant rate C due to friction. If \(M_{c}\) and \(R_{c}\) denote the mass and radius of the earth respectively, show that the satellite falls on the earth in a limit timet given by

\[t = \frac{GmM_{e}}{2C}\left( \frac{1}{R_{e}} - \frac{1}{r} \right)\]

Sol. Let velocity of satellite in its orbit of radius r be \(v\) then we have

\[v = \sqrt{\frac{GM_{e}}{r}}\]

When satellite approaches earth's surface, if its velocity becomes \(v^{'}\) then it is given as

\[v^{'} = \sqrt{\frac{GM_{e}}{R_{e}}}\]

The total initial energy of satellite at a distance \(r\) is

\[\begin{matrix} & E_{T_{i}} = K_{i} + U_{i} \\ & \ = \frac{1}{2}{mv}^{2} - \frac{{GM}_{e}\text{ }m}{r} \\ & \ = - \frac{1}{2}\frac{{GM}_{e}\text{ }m}{r} \end{matrix}\]

The total final energy of satellite at a distance \(R_{e}\) is

\[\begin{matrix} & {EE}_{T_{f}} = K_{f} + U_{f} \\ & \ = \frac{1}{2}{mv}^{'2} - \frac{{GM}_{c}\text{ }m}{R_{e}} \end{matrix}\]

\[= \frac{1}{2}\frac{{GM}_{e}\text{ }m}{R_{e}}\]

As satellite is loosing energy at a rate \(C\), if it take a time \(t\) in reaching earth, we have

\[\begin{matrix} & Ct = E_{T_{f}} - E_{T_{f}} \\ & \ = \frac{1}{2}{GM}_{e}\text{ }m\left\lbrack \frac{1}{R_{e}} - \frac{1}{r} \right\rbrack \\ & t = \frac{{GM}_{e}\text{ }m}{2C}\left\lbrack \frac{1}{R_{e}} - \frac{1}{r} \right\rbrack \end{matrix}\]

or
Q. 8 Two Earth's satellites move in a common plane along circular orbits. The orbital radius of one satellite \(r = 7000\text{ }km\) while that of the other satellite is \(\Delta r = 70\text{ }km\) less. What time interval separates the periodic approaches of the satellites to each other over the minimum distance?

Sol. Now for first satellite which is revolving about the earth (mass M and radius r ) the orbital speed is

\[v = \sqrt{\frac{GM}{r}}\]

Let \(T_{1}\) and \(T_{2}\) be the time period for first and second satellites respectively. Then we know that

\[T_{1} = \frac{2\pi r}{v} = \frac{2\pi r}{\sqrt{GM}}r^{3/2}\]

and

\[T_{2} = \frac{2\pi}{\sqrt{GM}}(r - \Delta r)^{3/2}\]

As second satellite is revolving in a radius \((r - \Delta r)\). Know the period interval \(\left( T_{1} - T_{2} \right)\) is given by

\[\begin{matrix} T_{1} - T_{2} & \ = \frac{2\pi}{\sqrt{GM}}\left\lbrack r^{3/2} - r^{3/2}\left( 1 - \frac{\Delta r}{r} \right)^{3/2} \right\rbrack \\ & \ = \frac{2\pi}{\sqrt{GM}}r^{3/2}\left\lbrack 1 - \left( 1 - \frac{3}{2}\frac{\Delta r}{r} \right) \right\rbrack \\ & \ = \frac{2\pi}{\sqrt{GM}}r^{3/2}\left( \frac{3}{2}\frac{\Delta r}{r} \right) \end{matrix}\]