Question

Acceleration due to gravity on surface of earth is 10 m/s^2 and radius of earth is 6400 km. with what minimum velocity must a body be thrown from the surface of earth so that it reaches a height of 6400 km ?
(Please provide steps)
A. 8 km/s
B. 64 km/s
C. 1 km/s
D. 32 km/s

Answer 1

we know ,
according to Gravitational concept ,
g =g°/( 1+ h/r)²

g varies with h ,

g is what ?? this is acceleration ,
so, g = vdv/dh

use this here,
vdv/dh =g°/(1+h/r)²
vdv = g° .dh /{ 1/( 1+h/r)²
integrate both sides
[v²/2 ]= g°r²[ { -1/( r + h)} ] put limit
( v²/2 -0) = -g°r²{ 1/( r+ r) -1/(r + 0)}
v²/2 = g°r²/2r = g°r/2
v² = g°r
v =√(g°r)

given,
g° = 10 m/sec²
r = 6400 km
v=√(10×10-³ ×6400)
=√(64) =8 km/sec

hence option A is correct