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?
The weight of a body is the force acting one it due to earth’s gravity and is the product of mass and acceleration due to gravity is given as
W = mg
Where W is the weight of a body of mass m on the surface of the earth, g is acceleration due to gravity on the surface of the earth
Now as the height from the surface of the earth increases acceleration due to gravity decreases, so the weight of the body also decreases i.e.
Wh = mgh
Where Wh is the weight of a body of mass m kept on h height above the surface of the earth, Let gh be the acceleration due to gravity at a height h above the surface of the earth
Now acceleration due to gravity above surface h varies according to the relation
Here,
gh is the acceleration due to gravity at a height h above the surface of the earth, g is acceleration due to gravity on earth’s surface and R is Radius of the Earth
The situation has been shown in figure
In this case, height is equal to half of the radius of earth i.e.
H = R/2
Therefore,
On solving further we get
i.e. we get
gh = 4/9 g
So acceleration due to gravity at a height R/2 above the surface of the earth is 4/9 times to that on the surface of the earth
So weight at that height will be
Wh = m(4/9g) = 4/9 mg
i.e.
Wh = 4/9 W
So weight will be 4/9 times weight at the surface of the earth
We are given weight on the surface of the earth as
W = 63 N, so we have
Wh = 4/9 × 63N = 28 N
So the weight of the body or gravitational force on it due to the earth at a height equal to half the radius of the earth will be 28 N