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Q7.15:
A body weighs 63 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 ?

Solution :

Step 1: Identifying Given Variables and Physical Constants

According to the universal law of gravitation, the weight of an object of mass $m$ on the surface of the Earth (radius $R$) is given by the gravitational force $F_s$:

$F_s = \frac{GMm}{R^2} = 63\text{ N}$

Where:

  • $G$ = Universal Gravitational Constant
  • $M$ = Mass of the Earth
  • $R$ = Radius of the Earth
  • $h = \frac{R}{2}$ (The height at which we must calculate the force)

Step 2: Defining the Gravitational Force at Height $h$

The gravitational force $F_h$ at a distance $r$ from the center of the Earth is given by the inverse-square law [Per Newton's Law of Universal Gravitation]:

$F_h = \frac{GMm}{(R + h)^2}$

Step 3: Substituting and Simplifying the Algebraic Expression

Substitute $h = \frac{R}{2}$ into the denominator:

$F_h = \frac{GMm}{(R + \frac{R}{2})^2}$

$F_h = \frac{GMm}{(\frac{3R}{2})^2}$

$F_h = \frac{GMm}{\frac{9R^2}{4}}$

$F_h = \frac{4}{9} \left( \frac{GMm}{R^2} \right)$

Step 4: Calculation using Given Weight

Since we know $\frac{GMm}{R^2} = 63\text{ N}$ from Step 1, we substitute this value into the expression derived in Step 3:

$F_h = \frac{4}{9} \times 63\text{ N}$

$F_h = 4 \times 7\text{ N}$

$F_h = 28\text{ N}$

Visual Representation of Gravitational Potential Field

Earth R R/2 Body (28 N)

Final Solution: The gravitational force on the body at a height equal to half the radius of the earth is 28 N.


More Questions from Class 11 Physics Gravitation EXERCISES


CBSE Solutions for Class 11 Physics Gravitation


Chapters in CBSE - Class 11 Physics


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