Question

Derive the expressions for orbital velocity and time period of satellite.

Answer 1

A celestial body which revolves around a planet is called a satellite. Moon is the only natural satellite of the earth.

(i) Orbit speed of a satellite: Suppose a satellite of mass m is revolving around a planet of mass M and radius R, at height of h from the surface of the planet with orbital speed v0.
∴ Radius of the orbit of satellite
r = (R + h)
The gravitational attraction acting between the planet and the satellite provides the required centripetal force for circular motion of the satellite.
i.e.,
Centripetal force = Gravitational force between the planet and satellite

∵ r = (R + h)

If g we the value of gravitational acceleration on the surface of the planet, then

If the satellite is revolving near the surface of the planet then,

h << R \(\Rightarrow\frac{h}{r}<<1\)

Hence, \(\frac{h}{R}\) can be neglected.

\(\therefore v_o=\sqrt{gR}\) ........(4)

Calculation of orbital velocity of satellite revolving near the earth's surface:

\(\therefore v_o=\sqrt{gR}\)

For earth, Re = 6.4 x 10⁶ m; g = 9.8 m/s²

\(\therefore v_o=\sqrt{9.8\times6.4\times 10^6}\) \(=\sqrt{98\times64\times 10^4}\)

\(=\sqrt{2\times49\times64\times 10^4}\)\(=7\times8\times10^2\sqrt{2}\)

= 56 x 1.44 x 10²

= 7.9184 x 10³ m/s = 7.9 km s^-1

Period of revolution of the satellite: The time taken by satellite in completing one revolution around the planet, is called the time period of the satellite. It is denoted by T.

For satellite revolving near the surface of the planet,
h << R \(\Rightarrow\frac{h}{R}<<1\)

∴ On neglecting \(\frac{h}{R}\)

T=\(2\pi \sqrt{\frac{R}{g}}\)  ....(6)

Again from equation (5),

T=\(2\pi \sqrt{\frac{(R+h)^3}{GM}}\)

For satellite revolving near the surface of the planet

h << R

∴ On neglecting h, T = \(2\pi \sqrt{\frac{(R)^3}{GM}}\)

∴ M = \(\frac{4}{3}\pi R^3 \rho\)

Thus there are following three formula for time period of satellite revolving near the surface of the planet

Time period of satellite revolving near the Earth's surface

∵ Time period T = \(2\pi \sqrt{\frac{R_e}{g}}\)

∴ Re = 6400 km = 64 x 10⁶ m; g = 9.8 m/s²

∴  T = \(2\times3.14\sqrt{\frac{6.4\times10^6}{9.8}}\)

= 5067 seconds = 86.6 minute
or T = 84.6 min.