### Orbital Speed:

Suppose that a satellite of mass m orbits a planet in a circular orbit of radius r.

$ \displaystyle F_g = \frac{G M m}{r^2} $

If v_{o} is the orbital velocity of the satellite, then

$ \displaystyle \frac{m v_0^2}{r} = \frac{G M m}{r^2} $

$ \displaystyle v_0 = \sqrt{\frac{G M}{r}} $

If the altitude of the orbit is h , then,

$ \displaystyle v_0 = \sqrt{\frac{G M}{R + h}} $

Angular Speed $ \displaystyle \omega = \frac{v_0}{r} = \sqrt{\frac{G M}{r^3}} $

__Time Period of Revolution :__

The period of revolution $ \displaystyle T = \frac{2 \pi }{\omega} $

$ \displaystyle T = 2\pi \sqrt{\frac{r^3}{G M} }=2\pi \sqrt{\frac{(R+h)^3}{G M} } $

__Energy of Satellites:__

__Potential Energy :__

The Gravitational potential energy of the planet satellite system is

$ \displaystyle P.E = -\frac{G M m}{r} $

__Kinetic Energy :__

KE of a satellite in orbit is given by

$ \displaystyle K.E = \frac{1}{2}m v_0^2 ; where , v_0 = \sqrt{\frac{GM}{r}} $

$ \displaystyle K.E = \frac{G M m}{2r} $

__Total Energy :__

The total mechanical energy of the satellite is equal to the sum of its potential and kinetic energies.

Total energy $ \displaystyle E = -\frac{G M m}{r} + \frac{G M m}{2r} $

$ \displaystyle E = -\frac{G M m}{2r} $

The total energy is negative because the satellite is bound by the planet’s gravitational field.

__Angular Momentum :__

The angular momentum of the satellite is given by

L = m v_{o}r

Putting $ \displaystyle v_0 = \sqrt{\frac{GM}{r}} $

$ \displaystyle L = m \sqrt{G M r} $

__Points to Remember:__

(a) The KE of the satellite is equal in magnitude to its total energy and one-half of its potential energy.

(b) If we want to remove the satellite from its orbit to infinity we have to impart an additional energy equal in magnitude to its total energy, that is GMm/2r

Illustration : In the figure shown a planet of mass m moves in an elliptical orbit around the sun of mass M. Compare its speeds, linear and angular momenta at the points A and B.

Solution: Kepler’s Law of Areas is a consequence of the conservation of angular momentum of the planet about the sun. Since the central force $ \displaystyle \vec{F_g}= \frac{GMm}{r^2}\hat{r} $ passes through the sun, the torque of this force about the sun is zero. Therefore its angular momenta about the sun at A and B remain constant

⇒ L_{A} = L_{B}

⇒ mv_{1}r_{1} = mv_{2}r_{2}

⇒ $\large \frac{v_1}{v_2} = \frac{r_2}{r_1}$

Ratio of angular momenta

$\large \frac{L_1}{L_2} =\frac{m v_1 r_1}{m v_2 r_2} = 1 $

$\large \frac{P_1}{P_2} =\frac{m v_1 }{m v_2 } = \frac{r_2}{r_1}$

Illustration : The mean distance of Jupiter from the sun is nearly 5.2 times the corresponding distance between earth and sun. Using Kepler’s Law, find the period of revolution of Jupiter in its orbit.

Solution: According to Kepler’s Law of period T^{2} ∝ r^{3}

where T = period of revolution, r = mean distance from sun

$ \displaystyle \frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3} $

$ \displaystyle \frac{T_1}{T_2} = (\frac{r_1}{r_2})^{3/2} $

Let T_{2} be the period of revolution of earth around sun and r_{2} be the mean distance between earth and sun.

Given: (r_{1}/r_{2}) = 5.2 and T_{2} = 365 days (one year).

Hence,$ \displaystyle T_1 = T_2(\frac{r_1}{r_2})^{3/2} $

T_{1} = 1 (5.2)^{3/2} = 11.858 years.

Illustration : Two hypothetical planets 1 and 2 are moving in the same elliptical path as shown in the figure. If the planets are situated at minimum and maximum distance from the sun and one of the planets, say 1, has speed v, find the relative angular speed of the planets for the given situation

Solution:

According to Kepler’s Law of area (1/2) r^{2}ω = constant.

⇒ r_{1}^{2}ω_{1} = r_{2}^{2}ω_{2 }

⇒ ω_{2 }/ω_{1 } = ( r_{1}/ r_{2})^{2}

⇒ ω_{2 }/ω_{1 } = ( r/ 2r)^{2}

ω_{2 } = ω_{1 }/4

putting ω_{1} = v_{1 }/r_{1} = v/r

we obtain, ω_{2 } = v/4r

Since the direction of angular velocities ω_{1} and ω_{2} are same, the relative angular speed

ω_{r} = |ω_{1} – ω_{2}| = |v/r – v/4r|

⇒ ω_{r} = 3v/4r

__Weightlessness :__

Weightlessness is a term that is usually applied to a state where the ‘measured weight’ i.e. the normal reaction (R) on a body from the floor / ground becomes zero. This occurs when the body falls freely under gravity, in satellites or spacecrafts