Question
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A body weighs 72 N on the surface of the earth. What is the gravitational force on it due to the Earth at a height equal to half the radius of the earth?
A.)16 N
B.)28 N
C.)32 N
D.)72 N

Answer Verified Verified
Hint: A good approach would be to determine the acceleration due to gravity from the universal law of gravitation for when the body is at the surface of the earth, and then find the acceleration due to gravity at a height h using the same law but in terms of g at surface.

Formula used:
Acceleration due to gravity $g = \dfrac{GM}{R^2}$ where G is the gravitational constant. M is the mass of the body exerting force and R is the distance to the body.
Force acting on a body due to gravitational acceleration $F = mg$ where m is the mass of the body and g is the acceleration due to gravity

Complete answer:
Let us begin by looking at the first scenario:
On the surface of the Earth, i.e., at a height R (where R is the radius of the Earth), $F= mg = 72\;N$.
From the universal law of gravitation we get: $F = \dfrac{GMm}{R^2} \Rightarrow mg = \dfrac{GMm}{R^2} \Rightarrow g = \dfrac{GM}{R^2}$

Now, if the body is raised to a height of $h = \dfrac{R}{2}$ then the acceleration due to gravity at that height will be:
$g_h = \dfrac{GM}{\left(R + h\right)^2} = \dfrac{GM}{\left(R + \dfrac{R}{2}\right)^2} = \dfrac{4}{9}\left( \dfrac{GM}{R^2}\right) \Rightarrow g_h = \dfrac{4}{9}\;g$

Therefore, the gravitational force acting on the body at height h is found to be:
$F_h = mg_h = m \times \dfrac{4}{9} \times g = \dfrac{4}{9} \times mg = \dfrac{4}{9} \times 72 \Rightarrow F_h = 32\;N$
Hence, the correct choice would be 32 N.

So, the correct answer is “Option C”.

Note:
Remember that the weight of the body is basically the gravitational force acting on it, which is why it is safe to assume mg = 72 N.
When a body is raised to a certain height from the surface of the earth, do not forget to include the radius of the earth to the distance since the point of origin of this acceleration due to gravity is assumed to be the planetary centre. However, when $h << R$ where R is the earth radius then you can use the approximation : $g_h = \dfrac{g}{1-\dfrac{2h}{R}}$.