The approximate height from the surface of the earth at which the weight of the body becomes \(\frac{1}{3}\) of its weight on the surface of earth is [Radius of earth R = 6400 km and √3=1.732]
- 3840 km
- 4685 km
- 2133 km
- 4267 km
The Correct Option is B
Solution and Explanation
According to the given information
\(\frac{GM}{(R+h)^2}\)=\(\frac{1}{3}\)×\(\frac{GM}{R^2}\)
⇒ R+h=√3 R
⇒ h=(√3-1)R∼4685 km
The correct option is (B) : 4685 km
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Concepts Used:
Gravitational Potential Energy
The work which a body needs to do, against the force of gravity, in order to bring that body into a particular space is called Gravitational potential energy. The stored is the result of the gravitational attraction of the Earth for the object. The GPE of the massive ball of a demolition machine depends on two variables - the mass of the ball and the height to which it is raised. There is a direct relation between GPE and the mass of an object. More massive objects have greater GPE. Also, there is a direct relation between GPE and the height of an object. The higher that an object is elevated, the greater the GPE. The relationship is expressed in the following manner:
PEgrav = mass x g x height
PEgrav = m x g x h
Where,
m is the mass of the object,
h is the height of the object
g is the gravitational field strength (9.8 N/kg on Earth) - sometimes referred to as the acceleration of gravity.