Question 319: Gravitation

Question

A body is projected vertically from earth's surface with velocity equal to half the escape velocity. The maximum height reached by the satellite is ( $$R$$ = radius of earth)

Accepted Answer

Given: $$v=\frac{v_e}{2}$$
If body is projected with velocity $$\mathrm{v}\left(\mathrm{v} < \mathrm{v}_{\mathrm{e}}ight)$$ then height up to which it will rise, $$h=\frac{R}{\left(\frac{v_c^2}{v^2}-1ight)}$$
$$	herefore \quad \mathrm{h}=\frac{\mathrm{R}}{\left(\frac{\mathrm{v}_{\mathrm{e}}}{\mathrm{v}_{\mathrm{e}} / 2}ight)^2-1}=\frac{\mathrm{R}}{4-1}=\frac{\mathrm{R}}{3}$$

Answer

$$\mathrm{R}$$

Answer

$$\frac{R}{2}$$

Answer

$$\frac{R}{3}$$

Answer

$$\frac{\mathrm{R}}{4}$$

Gravitation

A body is projected vertically from earth's surface with velocity equal to half the escape velocity. The maximum height reached by the satellite is ( $R$ = radius of earth)

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