Question 316: Gravitation

Question

The magnitude of gravitational field at distance ' $r_1$ ' and ' $r_2$ ' from the centre of a uniform sphere of radius ' $R$ ' and mass ' $M$ ' are ' $F_1$ ' and ' $F_2$ ' respectively. The ratio ' $\left(F_1 / F_2\right)$ ' will be (if $r_1>R$ and $r_2

Accepted Answer

To find the ratio $$\left(\frac{F_1}{F_2}ight)$$, we first calculate the gravitational field at $$r_1$$ and $$r_2$$.
For a point outside the uniform sphere ($r_1 > R$), the gravitational field $$F_1$$ is given by:
$$ F_1 = \frac{G M}{r_1^2} $$
For a point inside the sphere ($r_2 < R$), the gravitational field $$F_2$$ is given by:
$$ F_2 = \frac{G M_{	ext{enclosed}}}{r_2^2} $$
where the enclosed mass $$M_{	ext{enclosed}}$$ for a solid sphere of uniform density is given by
$$ M_{	ext{enclosed}} = M \left(\frac{r_2^3}{R^3}ight) $$
because the mass is proportional to the volume. Therefore:
$$ F_2 = \frac{G M \left(\frac{r_2^3}{R^3}ight)}{r_2^2} = \frac{G M r_2}{R^3} $$
Now, we find the ratio $$\left(\frac{F_1}{F_2}ight)$$:
$$ \frac{F_1}{F_2} = \frac{\frac{G M}{r_1^2}}{\frac{G M r_2}{R^3}} $$
Simplifying this expression, we have:
$$ \frac{F_1}{F_2} = \frac{R^3}{r_1^2 r_2} $$
Hence, the correct option is
Option C:
$$\frac{\mathrm{R}^3}{\mathrm{r}_1^2 \mathrm{r}_2}$$

Answer

$\mathrm{\frac{R^2}{r_1 r_2}}$

Answer

$\frac{\mathrm{R}^3}{\mathrm{r}_1 \mathrm{r}_2^2}$

Answer

$\frac{\mathrm{R}^3}{\mathrm{r}_1^2 \mathrm{r}_2}$

Answer

$\frac{\mathrm{R}^4}{\mathrm{r}_1^2 \mathrm{r}_2^2}$

Gravitation

The magnitude of gravitational field at distance ' r_1 ' and ' r_2 ' from the centre of a uniform sphere of radius ' R ' and mass ' M ' are ' F_1 ' and ' F_2 ' respectively. The ratio ' \left(F_1 / F_2\right) ' will be (if r_1>R and $r_2

Choose the correct answer: