Question 222: Gravitation

Question

If $$g$$ is the acceleration due to gravity on the earth's surface, the gain in the potential energy of an object of mass $$m$$ raised from the surface of the earth to a height equal to the radius $$R$$ of the earth is

Accepted Answer

Gravitational potential energy on the earth surface of a body
U = $$-{{GmM} \over R}$$
And at the height h from the earth surface the potential energy
$${U_h} =  - {{GmM} \over {R + h}}$$ = $$ - {{GmM} \over {2R}}$$  [ as h = R ]
So the gain in the potential energy
$$\Delta U = {U_h} - U$$
$$	herefore$$ $$\Delta U = {{ - GmM} \over {2R}} + {{GmM} \over R};$$
$$ \Rightarrow $$ $$\Delta U = {{GmM} \over {2R}}$$
Now $${{GM} \over {{R^2}}} = g;$$ $$\,\,\,$$ $$	herefore$$ $${\mkern 1mu} {{GM} \over R} = gR$$
$$	herefore$$ $$\Delta U = {1 \over 2}mgR$$

Answer

$${1 \over 4}mgR$$

Answer

$$2mgR$$

Answer

$${1 \over 2}mgR$$

Answer

$$mgR$$

Gravitation

If $g is the acceleration due to gravity on the earth's surface, the gain in the potential energy of an object of mass m raised from the surface of the earth to a height equal to the radius R$ of the earth is

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