Problem 8 - A Complete Resource Book in Physics for JEE Main

Question

At a height above the surface of the earth equal to the radius of the earth the value of \(g\) (acceleration due to gravity on the surface of the earth) will be nearly (A) Zero (B) \(\sqrt{g}\) (C) \(\frac{g}{4}\) (D) \(\frac{g}{2}\)

Step-by-Step Solution

Answer

The value of \(g'\) at a height equal to the Earth's radius is \(\frac{g}{4}\). Therefore, the correct answer is (C) \(\frac{g}{4}\).

Step 1: Understand the given problem and variables

We are given the value of g, which is the acceleration due to gravity on the surface of Earth. We have to find the value of g' at a height h equal to the Earth's radius R.

Step 2: Use the formula for universal law of gravitation

The formula for universal law of gravitation is given by: \[F = \frac{G M m}{(R + h)^2}\] Where F is the gravitational force, G is the gravitational constant, M is the mass of the Earth, m is the mass of the object, R is the Earth's radius, and h is the height above the Earth's surface.

Step 3: Relate gravitational force to acceleration

We know that \(F = m g'\). Therefore, we can write the formula for gravitational force in terms of g': \[m g' = \frac{G M m}{(R+h)^2}\]

Step 4: Find the expression for g'

From step 3, we can find the expression for g' by eliminating the mass of the object (m) and solving for g': \[g' = \frac{G M}{(R+h)^2}\]

Step 5: Substitute the value of h and compare with the value of g

We are given that h = R. So, \[g' = \frac{G M}{(2R)^2}\] Also, we have the expression for g when height h = 0, which is: \[g = \frac{G M}{R^2}\] Now let's divide g' by g to compare their values: \[\frac{g'}{g} = \frac{\frac{G M}{(2R)^2}}{\frac{G M}{R^2}} = \frac{1}{4}\]

Step 6: Arrive at the final answer

So, the value of \(g'\) at a height equal to the Earth's radius is \(\frac{g}{4}\). Therefore, the correct answer is (C) \(\frac{g}{4}\).