**Newton’s Law Of Universal Gravitation**

Any two particles in the universe attract each other. This force is called the **force of gravitation**.

This concept was given by Newton.

According to newton, “Any two bodies in universe attract each other with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them.”

**Mathematical expression:**

Lat A and B be two particle of mass m_{1} and m_{2} respectively. Let the distance AB = r.

By the law of gravitational, the particle A attracts the particle B with a force F such that

F ∝ m_{1}m_{2} (for a given pair of particles)

\( \text{F}\propto \frac{\text{1}}{{{\text{r}}^{\text{2}}}}\text{ }\left( \text{for a given separation between the particles} \right) \)

\( \text{So F}\propto \frac{{{\text{m}}_{\text{1}}}{{\text{m}}_{\text{2}}}}{{{\text{r}}^{\text{2}}}} \)

\( \text{or F}=\text{G}\frac{{{\text{m}}_{\text{1}}}{{\text{m}}_{\text{2}}}}{{{\text{r}}^{\text{2}}}} \)

Here G is a constant known as the universal constant of gravitation. G = 6.67 × 10^{-11} Nm^{2}/kg^{2}

G is independent of the masses of the bodies and the distance between them.

Newton’s law of gravitation is applicable for everybody in the universe.

**Note:** The force between any two bodies in the universe is called the force of gravitation while the force with which earth attracts a body is called the force of gravity.

**Some Scientific Phenomenon Based on Gravitational Force:**

- The gravitational force between the sun and the earth keeps the earth moving around the sun.
- The gravitational force b/w the earth and the moon keeps the moon moving around the earth.

Existence of our solar system is due to gravitational force. - Gravitation force of the sun and the moon on the earth’s water surface is responsible for the tides in sea.
- Atmosphere above the earth is held due to gravitational force of the earth.
- Gravitational force between the sun and planet keeps the planet moving around the sun.
- Gravitational force is responsible for providing the centripetal force required by the planets.
- The attractive force of the earth is responsible for providing the centripetal force required by moon.

**Newton’s Third Law of Motion and Law of Gravitation:**

- Newton’s third law of motion is applicable to gravitation also.

Example: If the earth exerts a force of attraction on a body, the body also exerts an equal and opposite force of attraction on the earth. - As a = F/m

mass of the body is larger, acceleration produced will be smaller and vice versa.

**Newton’s Law Of Gravitation Example Problems With Solutions**

**Example 1:** Calculate the force between two masses of 100 kg and 1000 kg separated by a distance of 10 m (G = 6.67 × 10^{-11} Nm^{2}/kg^{2}).

**Solution: ** According to Newton’s law of gravitation, force of attraction between two bodies is

\(F=\text{G}\frac{{{\text{m}}_{\text{1}}}{{\text{m}}_{\text{2}}}}{{{\text{r}}^{\text{2}}}} \)

Here, m_{1} = 100 kg; m_{2} = 1000 kg;

r = 10 m; G = 6.67 × 10^{-11 }Nm^{2}/kg^{2}

\(\therefore F=\frac{6.67\times {{10}^{-11}}\times 100\times 1000}{{{(10)}^{2}}}\)

= 6.67 × 10^{-8} N

**Example 2: **Given mass of earth = 6 × 10^{24} kg, radius of earth = 6.4 × 10^{6} m. Calculate the force of attraction experienced by a man of mass 50 kg.

**Solution: **Force of gravitation is given by the expression,

\(F=\text{G}\frac{{{\text{m}}_{\text{1}}}{{\text{m}}_{\text{2}}}}{{{\text{r}}^{\text{2}}}} \)

Here, mass of earth, m_{1} = 6 × 10^{24} kg;

mass of man, m_{2} = 50 kg

Distance between them is to be taken equal to the radius of earth.

∴ r = 6.4 × 10^{6} m

Substituting these values, we get

\(\text{F}=\frac{6.67\times {{10}^{-11}}\times 6\times {{10}^{24}}\times 50}{{{(6.4\times {{10}^{6}})}^{2}}}\)

F = 488.5 N

**Example 3: **Compare the gravitational forces exerted by the sun and the moon on earth. Which exerts a greater force on earth ?

(Given: mass of sun, M_{s} = 4 × 10^{31} kg; mass of moon, M_{m} = 6.3 × 10^{22} kg; distance between sun and earth, r_{se} = 1.3 × 10^{12} m and the distance between moon and earth, r_{me} = 4.5 × 10^{8}m)

**Solution: **If mass of sun is M_{s} and mass of earth is M_{e} and distance between the sun and earth is r_{se}, then force exerted by the sun on earth is

\( {{F}_{s}}=\frac{G{{M}_{s}}{{M}_{e}}}{{{({{r}_{se}})}^{2}}}\text{ }….\text{(1)} \)

Similarly, if mass of moon is M_{m}, mass of earth is M_{e}, the distance between moon and earth is r_{me}, then force exerted by moon on the earth is

\( {{F}_{m}}=\frac{G{{M}_{m}}{{M}_{e}}}{{{({{r}_{me}})}^{2}}}\text{ }….\text{(2)} \)

Dividing equation (1) by equation (2), we get

\( \frac{{{F}_{s}}}{{{F}_{m}}}=\frac{G{{M}_{s}}{{M}_{e}}}{{{({{r}_{se}})}^{2}}}\times \frac{r_{me}^{2}}{G{{M}_{m}}{{M}_{e}}} \)

\( =\frac{{{M}_{s}}}{{{M}_{m}}}\times \frac{{{({{r}_{me}})}^{2}}}{{{({{r}_{se}})}^{2}}} \)

\( =\frac{4\times {{10}^{31}}}{6.3\times {{10}^{22}}}\times {{\left( \frac{4.5\times {{10}^{8}}}{1.3\times {{10}^{12}}} \right)}^{2}} \)

= 76.07

∴ The force exerted by the sun on earth is about 76 times the force exerted by the moon on earth.

**Example 4: **If mass and radius of earth is 6.0 × 10^{24} kg and 6.4 × 10^{6} m respectively, calculate the force exerted by earth on a body of mass 1 kg. Also, calculate

(i) acceleration produced in the body of mass 1 kg, and

(ii) acceleration produced in the earth

**Solution: **From Newton’s law of gravitation, we know that the force of attraction between two bodies is given by

\(F=\text{G}\frac{{{\text{m}}_{\text{1}}}{{\text{m}}_{\text{2}}}}{{{\text{r}}^{\text{2}}}} \)

Here, m_{1} = mass of earth = 6.0 × 10^{24} kg;

m_{2} = mass of body = 1 kg

r = distance between the two bodies

= radius of earth = 6.4 × 10^{6} m

G = 6.67 × 10–11 Nm^{2}/kg^{2}

\( \therefore F=\frac{6.67\times {{10}^{-11}}\times 6.0\times \,{{10}^{24}}\times 1}{{{(6.4\times {{10}^{6}})}^{2}}} \)

= 9.8 N

This shows that earth exerts a force of 9.8 N on a body of mass of 1 kg. The body will exert an equal force of attraction of 9.8 N on earth.

**(i)** Acceleration produced in the body of mass 1kg

Force = mass × acceleration

∴ Acceleration,

\( a=\frac{F}{m}=\frac{9.8}{1}=9.8\text{ }m/{{s}^{2}} \)

Thus, the acceleration produced in a body of mass 1 kg due to attraction of earth is 9.8 m/s^{2} , which is quite large. Thus, when a body is released, it falls towards the earth with an acceleration of 9.8 m/s^{2} , which can be easily observed.

**(ii)** Acceleration produced in the earth

Similarly, acceleration of earth is given by

\( =\frac{\text{Force}\,}{\text{Mass}\,\text{of}\,\text{earth}}=\frac{9.8}{6.0\times {{10}^{24}}} \)

= 1.63 × 10^{-24} m/s^{2}

This shows that the acceleration produced in the earth by a body of mass 1 kg is 1.63 × 10^{-24} m/s^{2} which is very small and cannot be observed.