Question

52. The acceleration due to gravity on the surface of earth
is 10 m/s2 and the radius of earth is 6400 km. With what
minimum velocity must a body be thrown from the surface
of the earth so that is reaches a height of 6400 km?
(1) 8 km/s
(2) 64 km/s
(3) 1 km/s
(4) 32 km/s​

Answer 1

Answer:

Explanation:

g= 10 ms^-2

R= 6400 km

vdv/dh =g°/(1+h/r)²

vdv = g° .dh /{ 1/( 1+h/r)²

Integrating both sides we get;

[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)

v=√(10×10-³ ×6400)

=√(64) = 8 km/sec

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