Question 434: Gravitation

Question

The earth is assumed to be a sphere of radius R. A platform is arranged at a height R from the surface of the earth. The escape velocity of a body from this platform is fv,  where v is its escape velocity from the surface of the Earth. The value of f is

Accepted Answer

Escape velocity of the body from the surface of earth is v = $$\sqrt {2gR} $$
For escape velocity of the body from the platform potential energy + kinetic energy = 0
$$ - {{GMm} \over {2R}} + {1 \over 2}m{v^2} = 0$$
$$ \Rightarrow f{v_{escape}} = \sqrt {{{GM} \over {{R^2}}}.R}  = \sqrt {gR}  = fv$$
From the surface of the earth,  $${v_{escape}} = \sqrt {2gR} $$
$$ 	herefore f{v_{escape}} = {{{v_{escape}}} \over {\sqrt 2 }}$$, $$ 	herefore $$ $$f = {1 \over {\sqrt 2 }}$$

Answer

1/2

Answer

$$\sqrt 2 $$

Answer

1/$$\sqrt 2 $$

Answer

1/3

Gravitation

The earth is assumed to be a sphere of radius R. A platform is arranged at a height R from the surface of the earth. The escape velocity of a body from this platform is fv, where v is its escape velocity from the surface of the Earth. The value of f is

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