A body weighs 63 N on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth ?
Given: The weight of the body on the surface of earth is 63 N and the height is equal to the half the radius of the earth.
The acceleration due to gravity above the earth’s surface is given as,
g ′ = g ( 1 + h R E ) 2
Where, the radius of earth is R E , the acceleration due to gravity is g , the height is h .
By substituting the given values in the above equation, we get
g ′ = g ( 1 + R e 2 R e ) 2 = g ( 1 + 1 2 ) 2 = 4 9 g
The weight of the body above the earth surface is given as,
W ′ = m g ′ = m ( 4 9 g ) = 4 9 ( m g ) = 4 9 W
Where, the weight of the body at the surface of earth is W .
By substituting the given value in the above equation, we get
W ′ = 4 9 × 63 = 28 N
Thus, the gravitational force on the body due to the earth is 28 N .
Given: The weight of the body on the surface of earth is
The acceleration due to gravity above the earth’s surface is given as,
Where, the radius of earth is
By substituting the given values in the above equation, we get
The weight of the body above the earth surface is given as,
Where, the weight of the body at the surface of earth is
By substituting the given value in the above equation, we get
Thus, the gravitational force on the body due to the earth is
