A rocket is fired vertically upwards with a speed of `upsilon (=5 km s^(-1))` from the surface of earth. It goes up to a height `h` before returning to earth. At height `h` a body is thrown from the rocket with speed `upsilon_(0)` in such away so that the body becomes a satellite of earth. Let the mass of the earth, `M = 6 xx 10^(24) kg`, mean radius of the earth, `R = 6.4 xx 10^(6)m, G = 6.67 xx 10^(-11) Nm^(2) kg^(-2) , g = 9.8 ms^(-2)`.
Answer the following questions:
If this satellite is to be taken at double of the present height from the surface of the earth, then the new time period of revolution is

Options

A. `9330 s`
B. `20080 s`
C. `11000 s`
D. `29400 s`

Answer

A

Explanation

Initial orbital radius,
`R_(1) = 6.4 xx 10^(6) + 1.6 xx 10^(6) = 8 xx 10^(6)m`
Time period of revolution `T_(1) = 7100 s`
New orbital radius,
`R_(2) = 8 xx 10^(6) + 1.6 xx 10^(6) = 9.6 xx 10^(6)m`
New tme period of revolution `T_(2) = ?`
`T_(2) = T_(1) ((R_(2))/(R_(1)))^(3//2) = 7100 ((9.6 xx 10^(6))/(8 xx 10^(6)))^(3//2)`
`= 7100 (1.2)^(3//2) ~~ 9330 s`