Home
Class 11
PHYSICS
Prove that the ratio of the rate of chan...

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)`

Text Solution

Verified by Experts

`implies` Gravitational acceleration at .r. `(r ge R_e)` from the centre of earth .
`g(r) = (GM_e)/r^2`
Differentiating w.r.t. to distance r,
`(dg(r))/(dr) = (-2GM_e)/r^3`
`= -(2GM_e)/((8R_e)R_(e)^2)[ :. (GM_e)/R_e^2="ge"]`
`:. [(dg(r))/(dr)]_(2R_e) =(-"ge")/(4R_e)`
=`[(dg(r))/(dr)]_(2Re)/("ge")=-1/(4R_e)` .
Promotional Banner

Topper's Solved these Questions

  • GRAVITATION

    KUMAR PRAKASHAN|Exercise Section - C Objective Questions (VSQs) (Answer the following questions In short :)|11 Videos

  • GRAVITATION

    KUMAR PRAKASHAN|Exercise Section - C Objective Questions (VSQs)|80 Videos

  • GRAVITATION

    KUMAR PRAKASHAN|Exercise Section - B ADDITIONAL EXERCISE|4 Videos

  • LAW OF MOTION

    KUMAR PRAKASHAN|Exercise (QUESTION PAPER) SECTION-D|1 Videos

Similar Questions

Explore conceptually related problems

Give the magnitude of g on the surface of earth.

The value of 'g' on earth surface depends :-

What is the value of acceleration due to gravity at a height equal to helf the radius of earth, from surface of earth ? [take g = 10 m//s^(2) on earth's surface]

At which place on the surface of the earth the value of g is maximum ? Gives its reasons.

A body weighs W newton at the surface of the earth. Its weight at a height equal to half the radius of the earth, will be

At which height from the surface of earth polar satellite revolving around the earth ?

Derive the equation of g at depth d below the surface of earth.

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 gravity at the surface 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 .......

The acceleration due to gravity at a height 1km above the earth is the same as at a depth d below the surface of earth. Then :