Variation of Gravity with Height – Rankers Physics

Acceleration Due to Gravity and its variation: Practice Problem & Solution

The height at which the weight of a body becomes 1/9th its weight on the surface of earth (radius of earth is R):
\(h = 3R\)
\(h = R\)
\(h = \frac{R}{2}\)
\(h = 2R\)

Solution Explained:

To solve this problem, we apply the core principles of Acceleration Due to Gravity and its variation. Understanding the underlying formula is key to arriving at the correct answer below:

Since acceleration due to gravity varies with height as \(g' = g \frac{R^2}{(R+h)^2}\), setting \(g' = g/9\) gives \(R+h = 3R\), which simplifies to \(h = 2R\).

Leave a Reply

Your email address will not be published. Required fields are marked *

Similar Practice Questions

Explore conceptually related problems to test your knowledge.

1. The radii of circular orbits of two satellites A and B of the earth, are $4R$ and $R$, respectively. If the speed of satellite A is $3V$, then the speed of satellite B will be: (2010 Pre)
2. The distance of two planets from the sun are $10^{13}\text{ m}$ and $10^{12}\text{ m}$ respectively. The ratio of time periods of the planets is: (1994, 88)
3. Assertion (A): Gravitational potential of earth at every place upon it is negative. Reason (R): Every body on earth is bound by the attraction of earth.  
4. The gravitational force of attraction between two bodies is \(F\) newtons. If the mass of each body and the distance between them are doubled, then the gravitational force between them in newton is
5. The density of a newly discovered planet is twice that of earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of the earth. If the radius of the earth is $R$, the radius of the planet would be: (2004)