Assuming the radius of the earth to be `6.5 xx 10^(6)` m. What is the time period T and speed of satellite for equatorial orbit at `1.4 xx 10^(3)` km above the surface of the earth.

Options

A. `6831 s and 7174 " ms"^(-1)`
B. `34155 s and 3204 " ms"^(-1)`
C. `6831 s and 2144 " ms"^(-1)`
D. `2431 s and 3514 "ms"^(-1)`

Answer

A

Explanation

`r=R_("earth")+h=(6.4xx10^(6)+14xx10^(6))"m"`
`rArr r=7.8 xx 10^(6)"m"`
`T=[(4pi^(2)r^(3))/(GM_("earth"))]=6831s`
Speed of satellite, `v=sqrt((GM_("earth"))/(r))=7174 " ms"^(-1)`.