Question 193: Gravitation

Question

The mass density of a planet of radius R varies with the distance r from its centre as
$$\rho $$(r) = $${\rho _0}\left( {1 - {{{r^2}} \over {{R^2}}}} \right)$$.
Then the gravitational field is maximum at :

Accepted Answer

dm = $$ho $$dv
$$\int {dm} $$ $$= \int {{ho _0}} 4\pi {r^2}dr$$
$$ \Rightarrow $$ M = $$ \int {{ho _0}} 4\pi {r^2}dr$$
Also E = $${{GM} \over {{r^2}}}$$
$$ \Rightarrow E{r^2} = 4\pi G\,\int\limits_0^r {{ho _0}} \left( {1 - {{{r^2}} \over {{R^2}}}} ight){r^2}dr$$
$$ \Rightarrow E = 4\pi G{ho _0}\left( {{{{r}} \over 3} - {{{r^3}} \over {5{R^2}}}} ight)$$
For maximum E, $${{dE} \over {dr}} = 0$$
$${1 \over 3} - {{3{r^2}} \over {5{R^2}}}$$ = 0
$$ 	herefore $$ $$r = \sqrt {{5 \over 9}} R$$

Answer

$$r = {1 \over {\sqrt 3 }}R$$

Answer

r = R

Answer

$$r = \sqrt {{3 \over 4}} R$$

Answer

$$r = \sqrt {{5 \over 9}} R$$

Gravitation

The mass density of a planet of radius R varies with the distance r from its centre as $\rho (r) = {\rho _0}\left( {1 - {{{r^2}} \over {{R^2}}}} \right)$. Then the gravitational field is maximum at :

Choose the correct answer: