Question

Solve the problem.

How much time a satellite in an orbit at height 35780 km above earth's surface would take, if the mass of the earth would have been four times its original mass?

Answer 1

 Given:
Height of the satellite, h = 35780 km
Let the original mass of Earth be M. Then its new mass will be 4M. 
Time taken by the satellite to revolved around the Earth's orbit is given as

  \[T = \frac{2\pi R}{v_c}\]

Now, v_c is given as

\[v_c = \sqrt{\frac{GM}{R + h}}\]

Thus, 

\[T = \frac{2\pi(R + h)}{\sqrt{\frac{GM}{R + h}}} = \frac{2\pi(R + h)\sqrt{R + h}}{\sqrt{GM}} . . . . . (i)\]

\[ \Rightarrow T \propto \frac{1}{\sqrt{M}} . . . . . (ii)\]

Thus from equation (ii) we see that when the mass of the Earth becomes 4 times, the time period of revolution of satellite should be halved.
i.e.   \[T_{4M} = \frac{T}{2}\]        .....(iii)
Now, h = 35780 km
M = \[6 \times {10}^{24} kg\]

R =  \[6 . 4 \times {10}^5 m\]

G = 6.67 × 10^-11 N m² /kg²
Putting the values of h, M and R in first, we get
T = 24 h
Using (iii), we get

\[T_{4M} = \frac{24}{2} = 12 h\]