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Standard XII

Physics

Satellites

A uniform rin...

Question

A uniform ring of mass m is lying at a distance $\sqrt{3}$ a from the centre of a sphere of 8mass M just over the sphere (where a is the radius of the ring as well as that of the msphere), Then magnitude of gravilational force between them is

\frac{G M m}{8 a^{2}}

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\frac{G M m}{\sqrt{3} a^{2}}

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\sqrt{3} \frac{G M m}{a^{2}}

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\sqrt{3} \frac{G M m}{8 a^{2}}

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Solution

The correct option is D $\sqrt{3} \frac{G M m}{8 a^{2}}$
The sphere for outside objects can e treated as point mass at a distance $\sqrt{3} a$ .

$F = \frac{G M m \sqrt{3} a}{(a^{2} + (\sqrt{3} a)^{2})^{\frac{3}{2}}}$

$F = \frac{\sqrt{3} G M m}{8 a^{2}}$

option D is correct.

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A uniform ring of mass m is lying at a distance √3 a from the centre of a sphere of 8mass M just over the sphere (where a is the radius of the ring as well as that of the msphere), Then magnitude of gravilational force between them is

The correct option is D √3GMm8a2The sphere for outside objects can e treated as point mass at a distance √3a. F=GMm√3a(a2+(√3a)2)32F=√3GMm8a2option D is correct.

A uniform ring of mass m is lying at a distance $\sqrt{3}$ a from the centre of a sphere of 8mass M just over the sphere (where a is the radius of the ring as well as that of the msphere), Then magnitude of gravilational force between them is

The correct option is D $\sqrt{3} \frac{G M m}{8 a^{2}}$
The sphere for outside objects can e treated as point mass at a distance $\sqrt{3} a$ .

$F = \frac{G M m \sqrt{3} a}{(a^{2} + (\sqrt{3} a)^{2})^{\frac{3}{2}}}$

$F = \frac{\sqrt{3} G M m}{8 a^{2}}$

option D is correct.