Question

A planet has twice the mass of earth and of identical size. What will be the height above the surface of the planet where its acceleration due to gravity reduces by 36% of its value on its surface?

Answer 1

Hint: Basically there are four fundamental forces in nature. They are gravitational force, electrostatic force, strong nuclear force, weak nuclear force. Here the force in action is gravitational force and it is a long range force and the weakest of the four fundamental forces.

Formula used:
$g = \dfrac{{GM}}{{{r^2}}}$

Complete step by step answer:
Gravitational force will be acting between any two masses at a particular distance. Since the magnitude of that force is very small nobody feels it. There are some similarities between the gravitational force and electrostatic force. gravitational force acts between the two masses whereas electrostatic force acts between the two charges. Both follow the inverse square law which means that the force will be inversely proportional to the square of distance between them.
So Newton discovered this gravitational force and due to this everybody on the earth is acted upon by the force due to the earth and this is what we call as the weight of the body.
We have the expression for the force between the earth of mass ‘M’ and the body of mass ‘m’ which are separated at the distance ‘r’ from the center of any planet. If we consider the surface of planet then $r = R$ i.e radius of planet
That expression is
$F = \dfrac{{GMm}}{{{r^2}}}$
Where ‘G’ is the gravitational constant
If we divide that force with mass of a body we will get acceleration due to gravity at that place so acceleration due to gravity on planet’s surface (g) will be
$g = \dfrac{{GM}}{{{r^2}}}$
The acceleration due to gravity on the surface of the planet is given by
$g = \dfrac{{GM}}{{{R^2}}}$
The acceleration due to gravity at any height ‘h’ from the surface of the planet is given by
${g_h} = g{\left[ {\dfrac{R}{{R + h}}} \right]^2}$
The condition given is at height ‘h’ gravity must be reduced by 36 percent that of the surface gravity. So
$\eqalign{
  & \Rightarrow {g_h} = g - \dfrac{{36g}}{{100}} \cr
  & \Rightarrow {g_h} = \dfrac{{64}}{{100}}g \cr
  & \Rightarrow g{\left[ {\dfrac{R}{{R + h}}} \right]^2} = \dfrac{{64}}{{100}}g \cr
  & \Rightarrow \dfrac{R}{{R + h}} = \dfrac{8}{{10}} \cr
  & \Rightarrow h = \dfrac{R}{4} \cr
  & \Rightarrow h = \dfrac{{6400}}{4} \cr
  & \therefore h = 1600km \cr} $
Since it is given that the planet has the same size as earth then its radius will be the same as the earth radium which is 6400km. so we substituted it above.
Hence at the height of 1600km the condition will be satisfied.

Note:
This is the formula to find out the gravity at the height ‘h’ from the surface of the planet. There will be a different formula to find out the gravity value at depth ‘h’ from the surface of the planet. The planet mass which is inside the circumference passing through that point only will contribute for the acceleration due to gravity. The outer mass won’t contribute.