Question

a satellite of 10 ton is increased into its orbit at a height of double the radius of earth from surface. what 'g' will it experience

Answer 1

it will experience 9.8m/s2

Answer 2

The answer should be 1.08ms^-2

As

at suface gs = GMe/Re^2 ------ (1.)

at distance 'd' in the sky gd = GMe/Re^2 = GMe/(3Re)^2 = GMe/9Re^2 --------- (2.)

2/1 = GMe/9Re^2÷GMe/Re^2 = GMe/9Re^2 × Re^/ GMe =

After Cuting we got

1/9

So, gd = gs/9 = 9.8/9 m.s^-2 = 1.08 m.s^-2

Hope it helpful