NCERT Solutions for Class 9 Science Chapter 10 Gravitation

Ans : The universal law of gravitation states that every object in the universe attracts every other object with a force called the gravitational force. The force acting between two objects is directly proportional to the product of their masses and inversely proportional to the square Of the distance between their centers.
For two objects of masses m1 and m2 and the distance between them r, the force (F) of attraction acting between them is given by the universal law of gravitation as:
\(F = \frac { \mathrm { G } m _ { 1 } m _ { 2 } } { r ^ { 2 } }\)
Where, G is the universal gravitation constant given by:
\(\mathrm { G } = 6.67 \times 10 ^ { - 11 } \mathrm { Nm } ^ { 2 } \mathrm { kg } ^ { - 2 }\) 

Ans : Let ME be the mass of the Earth and m be the mass of an object on its surface. If R is the radius of the Earth, then according to the universal law of gravitation, the gravitational force (F) acting between the Earth and the object is given by the relation:
\(F = \frac { G m _ { 1 } m _ { 2 } } { r ^ { 2 } }\) 

Ans : Gravity of the Earth attracts every object towards its centre. When an object is released from a height, it falls towards the surface of the Earth under the influence of gravitational force. The motion of the object is said to have free fall. 

Ans : When an object falls towards the ground from a height, then its velocity changes during the fall. This changing velocity produces acceleration in the object. This acceleration is known as acceleration due to gravity (g). Its value is given by 9.8 m/s2. 

Ans : Let ME be the mass of the Earth and m be an object on the surface of the Earth. Let RE be the radius of the Earth. According to the universal law of gravitation, weight WE of the object on the surface of the Earth is given by,
\(W _ { \mathrm { E } } = \frac { G M _ { \mathrm { E } } m } { R _ { \mathrm { E } } ^ { 2 } }\) Let MM and RM the mass and radius of the moon. Then, according to the universal law of gravitation, weight WN of the object on the surface of the moon is given by:
\(W _ { \mathrm { M } } = \frac { \mathrm { G } M _ { \mathrm { M } } m } { R _ { \mathrm { M } } ^ { 2 } }\)
\(\frac { W _ { \mathrm { M } } } { W _ { \mathrm { E } } } = \frac { M _ { \mathrm { M } } R _ { \mathrm { E } } ^ { 2 } } { M _ { \mathrm { E } } R _ { \mathrm { M } } ^ { 2 } }\)
\(M _ { \mathrm { E } } = 5.98 \times 10 ^ { 24 } \mathrm { kg } , M _ { \mathrm { M } } = 7.36 \times 10 ^ { 22 } \mathrm { kg }\)
\(R _ { \mathrm { E } } = 6.4 \times 10 ^ { 6 } \mathrm { m } , R _ { \mathrm { M } } = 1.74 \times 10 ^ { 6 } \mathrm { m }\)
 \(\therefore \frac { W _ { \mathrm { M } } } { W _ { \mathrm { E } } } = \frac { 7.36 \times 10 ^ { 22 } \times \left( 6.37 \times 10 ^ { 6 } \right) ^ { 2 } } { 5.98 \times 10 ^ { 24 } \times \left( 1.74 \times 10 ^ { 6 } \right) ^ { 2 } } = 0.165 \approx \frac { 1 } { 6 }\)
Therefore, weight  of an object on the moon is ⅙ of its weight on the Earth.