Name: Gravitational Potential Energy
Uploaded: 2019-07-19
Duration: 9 min 10 s
Description: The energy possessed by a body because of its position in a gravitational field is known as Gravitational Potential Energy. Let’s learn about it in detail in...

Question

Derive the time period of satellite orbiting the Earth.

Accepted Answer

The distance covered by the satellite during one rotation in its orbit is equal to $2 π (R_{e} + h)$ and time taken for it is the time period, T. Then
$S p e e d = \frac{D i s \tan c e t r a v e l l e d}{T i m e t a k e n} = \frac{2 π (R_{e} + h)}{T}$
$\sqrt{\frac{G M_{e}}{(R_{e} + h)}} = \frac{2 π (R_{e} + h)}{T} T = \frac{2 π {(R_{e} + h)}^{\frac{3}{2}}}{\sqrt{G M_{e}}}$
Squaring both sides of the equation , we get
$T^{2} = \frac{4 π^{2} {(R_{e} + h)}^{3}}{G M_{e}} \frac{4 π^{2}}{G M_{e}} = c (c o n s \tan t) T^{2} = c {(R_{e} + h)}^{3}$
above Equation implies that a satellite orbiting the Earth has the same relation between time and distance as that of Kepler’s law of planetary motion. For a satellite orbiting near the surface of the Earth, h is negligible compared to the radius of the Earth R_e. Then,
$T^{2} = \frac{4 π^{2} {R_{e}}^{3}}{G M_{e}} T^{2} = \frac{4 π^{2} R_{e}}{\frac{G M_{e}}{{R_{e}}^{2}}} T^{2} = \frac{4 π^{2} R_{e}}{g} \sin c e \frac{G M_{e}}{{R_{e}}^{2}} = g T = 2 π \sqrt{\frac{R_{e}}{g}}$

Planet (imaginary)	Time period (T) (in year)	Semi major axis (a) (in AU)
Kurinji	2	8
Mullai	3	18
Marutham	4	32
Neithal	5	50
Paalai	6	72

Derive the time period of satellite orbiting the Earth.

Important Questions on Gravitation

Learn and Prepare for any exam you want !