Question 139: Gravitation

Question

A body of mass $$\mathrm{m}$$ is projected with velocity $$\lambda \,v_{\mathrm{e}}$$ in vertically upward direction from the surface of the earth into space. It is given that $$v_{\mathrm{e}}$$ is escape velocity and $$\lambda<1$$. If air resistance is considered to be negligible, then the maximum height from the centre of earth, to which the body can go, will be :
(R : radius of earth)

Accepted Answer

Using energy conservation
$$ - {{G{M_e}m} \over {{R_e}}} + {1 \over 2}m{\left( {\lambda \sqrt {{{2G{M_e}} \over {{R_e}}}} } \right)^2} =  - {{G{M_e}m} \over r}$$
$${{G{M_e}m} \over r} = {{G{M_e}m} \over {{R_e}}} - {{G{M_e}m} \over {{R_e}}}{\lambda ^2}$$
$$r = {{{R_e}} \over {1 - {\lambda ^2}}}$$

Answer

$$\frac{\mathrm{R}}{1+\lambda^{2}}$$

Answer

$$\frac{R}{1-\lambda^{2}}$$

Answer

$$\frac{R}{1-\lambda}$$

Answer

$$\frac{\lambda^{2} \mathrm{R}}{1-\lambda^{2}}$$

Gravitation

Choose the correct answer: