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The energy required to take a satellite ...

The energy required to take a satellite to a height ‘h’ above Earth surface (radius of Earth`=6.4xx10^3` km ) is `E_1` and kinetic energy required for the satellite to be in a circular orbit at this height is `E_2`. The value of h for which `E_1` and `E_2` are equal, is:

A

`6.4 xx 10^(3) km`

B

`1.6 xx 10^(3) km`

C

`1.28 xx 10^(4) km`

D

`3.2 xx 10^(3) km`

Text Solution

Verified by Experts

The correct Answer is:
D
`epsilon_(1) = (-GMm)/((R +h)) + (GMm)/(R )`
`GMm((-R + R + h)/(R(R + h))`
`epsilon_(1) = (GMmh)/(R(R + h))`
`epsilon_(1) = (GMmh)/(R (R + h))" "……(i)`
`v = sqrt((GM)/(R + h))`
`KE = E_(2) = 1/2 (GM)/((R + h))`
`(GMmh)/(R(R + h)) = (GMm)/(2(R + h))`
`2h = R`
`h = 3.2 xx 10^(3) km`
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