Question 381: Gravitation

Question

A particle of mass 'm' is projected with a velocity $$\upsilon $$ = kV
e
= (k < 1) from the surface of the earth. (V
e
= escape velocity)
The maximum height above the surface reached by the particle is :

Accepted Answer

$$ - {{GMm} \over R} + {1 \over 2}m{k^2}{v_e}^2 =  - {{GMm} \over r}$$
$$ - {{GMm} \over R} + {1 \over 2}m{k^2}{{2GM} \over R} =  - {{GMm} \over r}$$
$$ - {1 \over R} + {{{k^2}} \over R} =  - {1 \over r}$$
$${1 \over r} = {1 \over R} - {{{k^2}} \over R}$$
$${1 \over r} = {{1 - {k^2}} \over R}$$
$$r = {R \over {1 - {k^2}}}$$
$$R + h = {R \over {1 - {k^2}}}$$
$$h = {R \over {1 - {k^2}}} - R = {{{k^2}} \over {1 - {k^2}}}R$$

Answer

$${{R{k^2}} \over {1 - {k^2}}}$$

Answer

$$R{\left( {{k \over {1 - {k^2}}}} \right)^2}$$

Answer

$$R{\left( {{k \over {1 + {k^2}}}} \right)^2}$$

Answer

$${{{R^2}k} \over {1 + k}}$$

Gravitation

A particle of mass 'm' is projected with a velocity $\upsilon $ = kV e = (k < 1) from the surface of the earth. (V e = escape velocity) The maximum height above the surface reached by the particle is :

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