Question

true or false. outside the earth ,g varies as 1/(R+h)square ​

Answer 1

The correct answer is true

Answer 2

Answer:

'outside the earth ,g varies as 1/(R+h)square' : It is a true statement. ​

Explanation:

• Outside the earth, Gravitational force g varies as 1/(R+h)^2, where `h` is the height above the surface of earth .

• Force F = $GMm^{}$/$R^{2}$, where G= Gravitational Constant , M= Mass of Earth, m = Mass of any object and R is the radius.

• Now,  F = m.g

• Hence, we can write, m.g = GMm/ R2

                            ⇒     g = GM/R2......................................(i)

• Now, if the element is at a height h, then the gravitational force at height h is g1 :

                            g1= GM/(R+h)2......................................(ii)

(ii) ÷ (i)  will give :

 ⇒ g1 = $gR^{2}$ /$(R+h)^{2}$......................................................(iii)

• Now gravitational force g and radius of the earth R , both are constant.

• So, from equation (iii), we can conclude that g1, that is the gravitational force at height h, varies as 1/$(R+h)^{2}$.