An artificial satellite of the earth is launched in circular orbit in the equatorial plane of the earth and the satellite is moving from west to east. With respect to a person on the equator, the satellite is completing one round trip in `24 h`. Mass of the earth is `M=6xx10^(24)kg.`For this situation the orbital radius of the satellite is

Options

A. `2.66xx10^(4) km`
B. `6400 km`
C. `36,000 km `
D. `29,600 km`

Answer

A

Explanation

Here time period of satellite w.r.t. observe on equator is `24 th` and the satellite is moving from west to east, so angular velocity of satellite wrt earth's axis of rotation (considered as fixed) is, `omega=(2pi)/(T_(s))+(2pi)/(T_(e))`, where `T_(S)` and `T_(e)` are time periode of satellite and earth, respectively,
`rArr omega=(pi)/6(hr^(-1))=1.45xx10^(-4) rad//s`
From, `v=sqrt((GM)/r)rArr romega=sqrt((GM)/r)`
`rArr r^(3//2)=(sqrt(GM))/(omega)`
`rArr r=2.66xx10^(7)m=2.66xx10^(4)km`.