Consider two solid spheres of radii $R_{1} = 1 m, R_{2} = 2 m$ and masses $M_{1}$ and $M_{2}$ , respectively. The gravitational field due to sphere (1) and (2) are shown. The value of $\frac{M _{1}}{M _{2}}$ is:

Question

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Answer 1

Let $E_{1}, E_{2}$ be the gravitational field intensities on the surface of the two spheres.

$E_{1} = \frac{G M _{1}}{R _{1}^{2}}$ and $E_{2} = \frac{G M _{2}}{R _{2}^{2}}$

In a solid sphere, the gravitational field varies linearly from the center to the surface and is proportional to $R^{- 2}$ outside the sphere.

From the figure, we have $E_{1} = 2$ at the surface of the first spher and $E_{2} = 3$ at the surface of the second sphere.

$∴ 2 = \frac{G M _{1}}{R _{1}^{2}} = \frac{G M _{1}}{1}, G M_{1} = 2$ ...(i)

$∴ 3 = \frac{G M _{2}}{R _{2}^{2}} = \frac{G M _{2}}{2 ^{2}}, G M_{2} = 12$ ...(ii)

Dividing (i) by (ii),
$\frac{M _{1}}{M _{2}} = \frac{2}{1 2} = \frac{1}{1 6}$

Answer 2

Let

E_{1}, E_{2}

be the gravitational field intensities on the surface of the two spheres.

E_{1} = \frac{G M _{1}}{R _{1}^{2}}

and

E_{2} = \frac{G M _{2}}{R _{2}^{2}}

In a solid sphere, the gravitational field varies linearly from the center to the surface and is proportional to

R^{- 2}

outside the sphere.

From the figure, we have

E_{1} = 2

at the surface of the first spher and

E_{2} = 3

at the surface of the second sphere.

∴ 2 = \frac{G M _{1}}{R _{1}^{2}} = \frac{G M _{1}}{1}, G M_{1} = 2

...(i)

∴ 3 = \frac{G M _{2}}{R _{2}^{2}} = \frac{G M _{2}}{2 ^{2}}, G M_{2} = 12

...(ii)

Dividing (i) by (ii),

\frac{M _{1}}{M _{2}} = \frac{2}{1 2} = \frac{1}{1 6}