Question

At what altitude (h) above the earth's surface would
the acceleration due to gravity be one fourth of its
value at the earth's surface?
(1) h = R
(2) h = 4R
(3) h = 2R
(4) h = 16R​

Answer 1

Answer:

h=R

Explanation:g'=g/4

By relation

g'=g/ (1+h/r)^2

g/4=g/ (1+h/r)^2

By solving we get

h=R

Answer 2

Answer:

(1) h = R

Explanation:

Given that

h= Height above the earth

R= Radius of earth

Lets mass of body which at height h =m

Mass of earth =M

We know that gravitational force given as

F= GmM/R² = m g

So gravity due to earth g

g =  GM/R²             -----------1

Lets take at h acceleration of body is one forth of g

g ' = g /4

g' =  GM/(R+h)²        

  g/ 4=  GM/(R+h)²                                            -----------2

From equation 1 and 2

g / 4 = g R² /(R+h)²

(R+h)² = 4 R²

R+ h = 2 R

h = R

So the option (1) is correct.

(1) h = R