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Chapter 7: Problem 8

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) \(\frac{g}{2}\)

Short Answer

Expert verified
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 by step solution

01

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.
02

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.
03

Relate gravitational force to acceleration

We know that \(F = m g'\). Therefore, we can write the formula for gravitational force in terms of g': \[m g' = \frac{G M m}{(R+h)^2}\]
04

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': \[g' = \frac{G M}{(R+h)^2}\]
05

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: \[\frac{g'}{g} = \frac{\frac{G M}{(2R)^2}}{\frac{G M}{R^2}} = \frac{1}{4}\]
06

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) \(\frac{g}{4}\).

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Most popular questions from this chapter

In 1783, John Mitchell noted that if a body having same density as that of the sun but radius 500 times that of the sun, magnitude of its escape velocity will be greater than \(c\), the speed of light. All the light emitted by such a body will return to it. He, thus, suggested the existence of a black hole. \(v=c=\sqrt{\frac{2 G M}{R}}\) suggests that a body of mass \(M\) will act as a black hole if its radius \(R\) is less than or equal to a certain critical radius. Karl Schwarzchild, in 1926 , derived the expression for the critical radius \(R_{S}\) called Schwarzchild radius. The surface of the sphere with radius \(R_{S}\) surrounding a black hole is called event horizon. Density of the sun is (A) \(14.1 \mathrm{~kg} \mathrm{~m}^{-3}\) (B) \(141.1 \mathrm{~kg} \mathrm{~m}^{-3}\) (C) \(1410 \mathrm{~kg} \mathrm{~m}^{-3}\) (D) None of these

The period of a satellite in a circular orbit of radius \(R\) is \(T\). The period of another satellite in a circular orbit of radius \(4 R\) is (A) \(4 T\) (B) \(T / 4\) (C) \(8 \mathrm{~T}\) (D) \(T / 8\)

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If the length of a simple pendulum is equal to the radius \(R\) of the earth, its time period will be (A) \(2 \pi \sqrt{R / g}\) (B) \(2 \pi \sqrt{R / 2 g}\) (C) \(2 \pi \sqrt{2 R / g}\) (D) \(\pi \sqrt{R / 2 g}\)
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