Question

A rocket is fired vertically upwards with a speed of `upsilon (=5 km s^(-1))` from the surface of earth. It goes up to a height `h` before returning to earth. At height `h` a body is thrown from the rocket with speed `upsilon_(0)` in such away so that the body becomes a satellite of earth. Let the mass of the earth, `M = 6 xx 10^(24) kg`, mean radius of the earth, `R = 6.4 xx 10^(6)m, G = 6.67 xx 10^(-11) Nm^(2) kg^(-2) , g = 9.8 ms^(-2)`.
Answer the following questions:
The value of `h` is

Answer 1

Let the rocket be fired with velocity `upsilon` from the surface of Earth and it reaches a height `h` from the surface of Earth where its velocity becomes zero.
Total energy of rocket at the surface of Earth `= K.E. + P.E. = (1)/(2) m upsilon^(2) + ((-GM m)/(R ))`
At the highest point, `upsilon = 0, K.E. = 0` and `P.E. = - (GM m)/((R + h))`
Total energy `= K.E. + P.E. = 0 + ((-GM m)/(R + h)) = - (GM m)/(R + h)`.
According to law of caonservation of energy.
`(1)/(2)m upsilon^(2) - (GM m)/(R ) = - (GM m)/((R + h))`
or `(1)/(2) upsilon^(2) = (GM)/(R ) - (GM)/((R + h)) = (g R^(2))/(R ) - (gR^(2))/((R + h)) = gR (1-(R)/(R + h)) = gR ((h)/(R + h))`
or `upsilon^(2) (R + h) = 2 g Rh` or `R upsilon^(2) = 2 gRh - upsilon^(2) h = (2gR - upsilon^(2))h`
or `h = (R upsilon^(2))/(2g R - upsilon^(2)) = ((6.4 xx 10^(6) )xx (5 xx 10^(3))^(2))/(2 xx 9.8 xx(6.4 xx 10^(6)) - (5 xx 10^(3))^(2)) = 1.6 xx 10^(6) m`