JEE Physics Gravitation Notes Set 07

Newton's Law of Gravitation

Any two point masses attract each other with a force along the line joining them. The force is proportional to the product of the masses and inversely proportional to the square of the distance: \( F = Gm_1m_2/r^2 \), where \( G = 6.67\times10^{-11}\ \text{N m}^2\text{kg}^{-2} \). The field strength at a distance r from a point mass M is \( g = GM/r^2 \). Forces from several masses add as vectors.

Acceleration Due to Gravity

1. At the surface of a planet of mass M and radius R: \( g = GM/R^2 \). This can be written as \( g = \tfrac43\pi G\rho R \).
2. At height h above the surface: \( g_h = g\left(\dfrac{R}{R+h}\right)^2 \), which is about \( g(1 - 2h/R) \) for small h.
3. At depth d below the surface: \( g_d = g(1 - d/R) \). At the centre g is zero.
4. Inside a uniform sphere g grows linearly with r. Outside it falls as \( 1/r^2 \). The maximum is at the surface.
5. Because of the earth's rotation, g is smaller at the equator than at the poles. A spring balance shows this change, a beam balance does not.

Gravitational Potential Energy and Potential

1. Potential energy of two masses: \( U = -Gm_1m_2/r \) (zero at infinity).
2. Potential due to a point mass or a spherical shell (outside): \( V = -GM/r \).
3. Inside a thin spherical shell the potential is constant, \( V = -GM/R \), and the field is zero.
4. Potential at the centre of a uniform solid sphere: \( V = -\tfrac32 GM/R \); field relation \( g = -dV/dr \).

Illustration 1. At what distance from the earth's centre is the gravitational field zero between the earth and the moon if the distance between their centres is D and \( M_e = 81M_m \)?
Answer:
Let the point be at x from the earth. \( \dfrac{GM_e}{x^2} = \dfrac{GM_m}{(D-x)^2} \).
\( \dfrac{x}{D-x} = 9 \), so \( x = 0.9D \).

Escape Velocity and Orbital Velocity

1. Escape velocity from the surface: \( v_e = \sqrt{2GM/R} = \sqrt{2gR} \), about 11.2 km/s for the earth. It does not depend on the mass or the direction of projection.
2. Orbital speed at radius r: \( v_o = \sqrt{GM/r} \), and for a satellite just above the surface \( v_o = \sqrt{gR} \), so \( v_e = \sqrt2\,v_o \).
3. A body projected with speed \( v \) less than \( v_e \) rises to a height found from energy conservation: \( \tfrac12mv^2 - GMm/R = -GMm/r_{max} \).
4. A black hole is an object compressed below the Schwarzschild radius \( R_s = 2GM/c^2 \).

Satellites

1. Time period: \( T = 2\pi\sqrt{r^3/GM} \), so \( T^2 \propto r^3 \). A geostationary satellite has T = 24 h and orbits about 36000 km above the equator.
2. Kinetic energy \( K = GMm/2r \), potential energy \( U = -GMm/r \), total energy \( E = -GMm/2r = -K \).
3. Work needed to raise a satellite from radius \( r_1 \) to \( r_2 \): \( W = \tfrac{GMm}{2}\left(\dfrac1{r_1}-\dfrac1{r_2}\right) \).
4. Objects inside a satellite are weightless, because both the satellite and objects accelerate at the same rate.

Kepler's Laws

1. Law of orbits: every planet moves in an ellipse with the sun at one focus.
2. Law of areas: the line joining planet and sun sweeps equal areas in equal times, so angular momentum is constant and \( v_1r_1 = v_2r_2 \) at the extreme points. The planet moves fastest at perihelion.
3. Law of periods: \( T^2 \propto a^3 \), where a is the semi-major axis. Total energy of the orbit is \( -GMm/2a \).

Illustration 2. A comet is at distances \( 1.6\times10^{12} \) m and \( 8\times10^{12} \) m from the sun at the nearest and farthest points. If its speed at the nearest point is 60 m/s, find the speed at the farthest point.
Answer:
Angular momentum is conserved: \( v_1r_1 = v_2r_2 \).
\( v_2 = 60\times1.6/8 = 12 \) m/s.