Home
Class 11
PHYSICS
Find the height over the earth's surface...

Find the height over the earth's surface at which the weight of a body becomes half of its value at the surface.

Text Solution

Verified by Experts

The correct Answer is:
A, B, D
Let the height be h
`:. (1/2)(GM)/R^2=(GM)/((R+h)^2)`
or `2R^2=(R+h)^2`
or `sqrt2R=R+h`
`h=(sqrt2-1)R`
Promotional Banner

Similar Questions

Explore conceptually related problems

The height a which the weight of a body becomes 1//16th its weight on the surface of earth (radius R ) is

At what height above the earth's surface the acceleration due to gravity will be 1/9 th of its value at the earth’s surface? Radius of earth is 6400 km.

At which height from the earth's surface does the acceleration due to gravity decrease by 75 % of its value at earth's surface ?

At what depth below the surface does the acceleration due to gravity becomes 70% of its value in the surface of earth ?

Prove that the ratio of the rate of change of g at a height equal to the earth's radius from the surface of the earth of the value of g at the surface of the earth is equal to (-1)/(4R_e)

Weight fo a body of a mass m decreases by 1% when it is raised to height h above the earth's surface. If the body is taken to depth h in a mine, change in its weight is

If g is the acceleration due to gravity on earth's surface, the gain of the potential energy of an object of mass m raised from the surface of the earth to a height equal to the radius R of the earth is

The earth is assumed to be a sphere of radius R. A platform is arranged at a height R from the surface of the earth. The escape velocity of a body from this platform is fu, where v is its escape velocity from the surface of the earth the value of f is ..........

At which height above earth's surface is the value of 'g' same as in a 100 km depp mine ?