How much time would a satellite in an orbit at a height of 35780 km above the earth's surface take to complete one revolution around the earth , if the mass of the earth were four times its original mass ?

Explanation

Given : (1) R (Earth) = 6400 km = ` 6.4 xx 10^(6) ` m , M(Earth) = ` 6 xx 10^(24) ` kg
` therefore M= 4M = 4 xx 6 xx 10^(24) ` kg ,
h = 35780 km = ` 3.578 xx 10^(7) m = 35.78 xx 10^(6)` m ,.
G = ` 6.67 xx 10^(-11) N.m^(2) // kg^(2) , T = ? `
The time that the satellite would take to complete one revolution around the earth , T ` (2 pi (R +h))/v_(c)`
Now, ` v_(c) = sqrt((GM)/(R+h))`
` T = (2 pi (R + h))/( sqrt(GM // (R+h))) = (2pi)/sqrt(GM) . (R + h)^(3//2)`
`(2pi (6.4 xx 10^(6)m + 35.78 xx 10^(6)m)^(3//2))/(sqrt(6.67 xx 10^(-11)N.m^(2)//kg^(2)xx 4xx6xx10^(24)kg))`
` 2 xx 3.1342 xx ((42.18)^(3) xx 10^(4))/( 6.67 xx 2.4) `s= Approx ` 4.303 xx 10^(4)` s
Approx `( 4.303 xx 10^(4))/3600 ` hours
= 11 hours 57 min 10 seconds = Approx 11.95 h