Question 299: Gravitation

Question

A body starts from rest from a distance $\mathrm{R}_0$ from the centre of the earth. The velocity acquired by the body when it reaches the surface of the earth will be ( $R=$ radius of earth, $M=$ mass of earth)

Accepted Answer

According to law of conservation of energy,
$$\frac{1}{2} \mathrm{mv}^2=-\frac{\mathrm{GMm}}{\mathrm{R}_0}-\left(\frac{\mathrm{GMm}}{\mathrm{R}}ight)=\mathrm{GMm}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}_0}ight)$$
$	herefore \quad \mathrm{V}^2=2 \mathrm{GM}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}_0}ight)$
$	herefore \quad$ The velocity acquired by the body when it reaches the surface of earth is:
$$\mathrm{v}=\sqrt{2 \mathrm{GM}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}_0}ight)}$$

Answer

$2 \mathrm{GM}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}_0}\right)$

Answer

$\sqrt{2 \mathrm{GM}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}_0}\right)}$

Answer

$\mathrm{GM}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}_0}\right)$

Answer

$2 \mathrm{GM} \sqrt{\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}_0}\right)}$

Gravitation

A body starts from rest from a distance \mathrm{R}_0 from the centre of the earth. The velocity acquired by the body when it reaches the surface of the earth will be ( R= radius of earth, M= mass of earth)

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