JEE Physics Gravitation Notes Set 02

Gravitational Potential Energy

The potential energy of two masses \( m_1, m_2 \) separated by r (zero at infinity) is
\[ U = -\frac{G m_1 m_2}{r} \]
It is negative because gravity is attractive: work must be done against the field to separate the bodies. Energy needed to take a mass m from the surface to infinity is \( GMm/R = mgR \).
Near the surface the potential energy difference between heights 0 and h reduces to \( mgh \) only when \( h \ll R \). The exact change from R to R + h is \( \frac{GMmh}{R(R+h)} \).

Escape Velocity

Escape velocity is the minimum speed that lets a body leave a planet's gravity and just reach infinity with zero speed. Setting kinetic energy equal to the magnitude of potential energy:
\[ v_e = \sqrt{\frac{2GM}{R}} = \sqrt{2gR} = R\sqrt{\frac{8\pi G\rho}{3}} \]
For the earth, v_e = 11.2 km/s; for the moon, about 2.4 km/s. Escape speed does not depend on the mass of the projectile or on the direction of projection. A body projected with speed v greater than v_e retains speed \( \sqrt{v^2 - v_e^2} \) at infinity. If a planet's escape speed is made equal to the speed of light, R = 2GM/c², which defines a black hole.

Orbital Motion of Satellites

For a satellite at distance \( r = R + h \) from the centre, gravity supplies the centripetal force:
Orbital speed: \( v_0 = \sqrt{\frac{GM}{r}} = R\sqrt{\frac{g}{R+h}} \); near the surface \( v_0 = \sqrt{gR} \approx 7.9 \) km/s
Time period: \( T = 2\pi\sqrt{\frac{r^3}{GM}} \); near the surface about 84 minutes
Energies: \( K = \frac{GMm}{2r},\ U = -\frac{GMm}{r},\ E = -\frac{GMm}{2r} = -K \)
Escape speed is \( \sqrt2 \) times orbital speed, so the extra speed needed to escape from a near-surface orbit is about 3.3 km/s.
Orbital speed and period do not depend on the satellite's mass. A lower orbit means higher speed; a satellite that loses energy to drag moves to a lower orbit and speeds up. Moving a satellite from radius r1 to r2 needs energy \( \frac{GMm}{2}\left(\frac{1}{r_1}-\frac{1}{r_2}\right) \).

Geostationary satellite: it moves west to east in the equatorial plane with period 24 hours, so it stays above one point on the earth. Its orbit radius is about 42,400 km from the centre, about 36,000 km above the surface. Polar satellites orbit at lower height (about 500 to 800 km) through the poles, scanning the whole earth as it rotates beneath them.

Weightlessness: in a freely falling satellite, the body and its surroundings fall with the same acceleration, so no contact force acts and the apparent weight is zero even though gravity is still present.

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 the sun and a planet sweeps equal areas in equal times, \( \frac{dA}{dt} = \frac{L}{2m} = \) constant. This is conservation of angular momentum, so the planet moves fastest at perihelion and slowest at aphelion: \( v_1 r_1 = v_2 r_2 \).
3. Law of periods: \( T^2 = \frac{4\pi^2}{GM}a^3 \), where a is the semi-major axis \( a = \frac{r_{min}+r_{max}}{2} \). A graph of \( T^2 \) against \( a^3 \) is a straight line of slope \( \frac{4\pi^2}{GM} \).
Total energy of a satellite in an elliptical orbit is \( -\frac{GMm}{2a} \) and stays constant, while kinetic and potential energies exchange.

Worked example 1. A planet has twice the mass and twice the radius of the earth. Find the ratio of its escape velocity to that of the earth.
Answer:
\( v_e \propto \sqrt{M/R} = \sqrt{2/2} = 1 \)
The ratio is 1 : 1.

Worked example 2. The orbital radius of a planet is 4 times that of the earth. Find its period in years.
Answer:
\( T \propto r^{3/2} = 4^{3/2} = 8 \)
Period = 8 years.