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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 time its original mass ?

Text Solution

Verified by Experts

Given : New Mass of Earth (M) ` = 4 xx (6 xx 10^(24))kg`
`= 24 xx 10^(24) kg`.
Altitude of the satellite (h) ` = 35780 km` .
` = 35780 xx 10^(3)m`
To find : Time required for revolution . (T) = ?
Formulae : `V = sqrt((GM)/(R+h))`
`T = (2pi (R+h))/(v)`
Solution : `V = sqrt((GM)/(R+h))`
` therefore V = sqrt((6.67 xx 10^(-11) xx 24 xx 10^(24))/(6.4 xx 10^(6) + 35780 xx 10^(3)))`
`therefore V = sqrt(37.59 xx 10^(6))`
` V = 6.160 km //s`
` T = (2 pi (R + h))/(v)`
` = ( 2 xx 3.14 xx 6400 + 35780)/(6.16)`
` = 43001. 68 sec`
` T = 11.94 hr ~ 12 ` hours .
`therefore` The satellite will take approximately 12 hours to revolve around earth.
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