Answer

Verified

19.5k+ views

**Hint:**The gravitational force is defined as the force of attraction or repulsion that acts between two bodies of similar or different masses. The acceleration due to gravity is also the result between the force acting between mass of an object and earth.

**Formula used:**The formula of the gravitational force between two masses is given by,

$F = \dfrac{{G{m_1}{m_2}}}{{{r^2}}}$

Where G is the universal gravitational constant ${m_1}$ and ${m_2}$ are the masses of the two bodies and r is the distance between the bodies.

**Complete step by step solution:**

Let a small element dM on the ring and sphere of mass m and calculate the force of gravitation on the sphere due to the ring.

Angle $\theta $ is between the force dF and the vertical line joining the centre of the ring and the sphere.

The angle $\theta $ can be calculated as follows,

\[ \Rightarrow \tan \theta = \dfrac{R}{{R\sqrt 3 }}\]

\[ \Rightarrow \tan \theta = \dfrac{1}{{\sqrt 3 }}\]

\[ \Rightarrow \theta = {\tan ^{ - 1}}\dfrac{1}{{\sqrt 3 }}\]

\[ \Rightarrow \theta = 30^\circ \].

The force in the x-direction will be cancelled out but the force in the y-direction remains.

The force due to dM is given by,

$ \Rightarrow dF = \dfrac{{Gm\left( {dM} \right)}}{{{{\left( {\sqrt {{{\left( {R\sqrt 3 } \right)}^2} + {R^2}} } \right)}^2}}}$

$ \Rightarrow dF = \dfrac{{Gm\left( {dM} \right)}}{{{{\left( {\sqrt {4{R^2}} } \right)}^2}}}$

$ \Rightarrow dF = \dfrac{{Gm\left( {dM} \right)}}{{4{R^2}}}$

Taking force in the y-direction we get,

$ \Rightarrow d{F_y} = \dfrac{{Gm\left( {dM} \right)}}{{4{R^2}}}\cos 30^\circ $

As the value of $\cos 30^\circ $ is equal to$\dfrac{{\sqrt 3 }}{2}$.

$ \Rightarrow d{F_y} = \dfrac{{Gm\left( {dM} \right)}}{{8{R^2}}} \times \sqrt 3 $

$ \Rightarrow d{F_y} = \dfrac{{Gm\left( {dM} \right)}}{{8{R^2}}} \times \sqrt 3 $

Now let us calculate the total force on the sphere due to the ring, integrate the value of the force due to the small element of mass dM of the ring,

$\int\limits_{dM} {dF} = \int\limits_{dM} {\dfrac{{\sqrt {3 \cdot } Gm\left( {dM} \right)}}{{8{R^2}}}} $

After integrating we get,

$F = \dfrac{{\sqrt {3 \cdot } G \cdot M \cdot m}}{{8{R^2}}}$

**The gravitational force on the sphere by the ring is equal to $F = \dfrac{{\sqrt {3 \cdot } G \cdot M \cdot m}}{{8{R^2}}}$.**

**Note:**There is force of attraction or repulsion between everybody present on earth or outside the earth. If a body possesses mass then it feels force due to gravitation. The gravitational force is always attractive irrespective of the mass and the distance between.

Recently Updated Pages

If a wire of resistance R is stretched to double of class 12 physics JEE_Main

Let f be a twice differentiable such that fleft x rightfleft class 11 maths JEE_Main

Find the points of intersection of the tangents at class 11 maths JEE_Main

For the two circles x2+y216 and x2+y22y0 there isare class 11 maths JEE_Main

The path difference between two waves for constructive class 11 physics JEE_MAIN

What is the difference between solvation and hydra class 11 chemistry JEE_Main

Other Pages

Differentiate between reversible and irreversible class 11 physics JEE_Main

Give one chemical test to distinguish between the following class 12 chemistry JEE_Main

Chlorobenzene is extremely less reactive towards a class 12 chemistry JEE_Main

Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main

Assertion The melting point Mn is more than that of class 11 chemistry JEE_Main

Electric field due to uniformly charged sphere class 12 physics JEE_Main