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Chapter 6: Problem 696

If the radius of earth is \(\mathrm{R}\) then height \({ }^{\prime} \mathrm{h}\) ' at which value of ' \(\mathrm{g}\) ' becomes one-fourth is (A) \(\mathrm{R} / 4\) (B) \(3 \mathrm{R} / 4\) (C) \(\mathrm{R}\) (D) \(\mathrm{R} / 8\)

Short Answer

Expert verified
The height at which the gravitational acceleration (g) becomes one-fourth of its value on Earth's surface is \(h = R\). So, the correct option is (C).

Step by step solution

01

1. Formula for gravitational force

We know that the formula for gravitational force is given by: \[F = G \frac{m \cdot M}{r^2}\] Where: * F is the gravitational force * G is the gravitational constant * m and M are the two masses * r is the distance between the centers of masses
02

2. Calculate gravitational acceleration on Earth's surface

The acceleration due to gravity on the Earth's surface, g, is related to the gravitational force by: \( F = m \cdot g\) Using the formula from the first step, we can find the gravitational acceleration on Earth's surface: \(g = \frac{GM_\oplus}{R^2}\) Here, Mₑ represents the mass of Earth.
03

3. Find the value of g which is one-fourth

Given that the value of g becomes one-fourth at height h, the new value of g can be represented as: \(g_{new} = \frac{1}{4} \cdot g = \frac{1}{4}\frac{GM_\oplus}{R^2}\)
04

4. Calculate g at height h

We need to find the value of g at height h. The formula for g at height h can be given by: \(g_{h} = \frac{GM_\oplus}{(R+h)^2}\)
05

5. Equate g_new with g_h and solve for h

Now, equate g_new with g_h: \(\frac{1}{4}\frac{GM_\oplus}{R^2} = \frac{GM_\oplus}{(R+h)^2}\) Solve for h: \(\frac{1}{4}R^2 = (R+h)^2\) \(R^2 = 4(R+h)^2\) \(R^2 = 4R^2 + 8Rh + 4h^2\) \(-3R^2 = 8Rh + 4h^2\) \(-\frac{3}{4}R^2 = 2h^2 + 4Rh\) \(-\frac{3}{8}R^2 = h^2 + 2Rh\) Now, we have a quadratic equation in terms of h: \(h^2 + 2Rh -\frac{3}{8}R^2 = 0\)
06

6. Solve the quadratic equation

Now, solve this quadratic equation for h using the quadratic formula: \(h = \frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\) In this case, we have \(a = 1\), \(b = 2R\), and \(c = -\frac{3}{8}R^2\). Plug in the values and solve for h: \(h = \frac{-2R\pm\sqrt{(2R)^2 - 4(\frac{3}{8}R^2)}}{2}\) \(h = -R\pm\sqrt{R^2}\) One solution is \(h = -R\), but the height cannot be negative, so we can discard this solution. The other solution is \(h = R\).
07

Conclusion

Therefore, the height at which the gravitational acceleration (g) becomes one-fourth of its value on Earth's surface is \(h = R\). So, the correct option is (C).

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Most popular questions from this chapter

When a particle is projected from the surface of earth, it mechanical energy and angular momentum about center of earth at all time is constant (i) A particle of mass \(\mathrm{m}\) is projected from the surface of earth with velocity \(\mathrm{V}_{0}\) at angle \(\theta\) with horizontal suppose \(\mathrm{h}\) be the maximum height of particle from surface of earth and \(\mathrm{v}\) its speed at that point them \(\mathrm{V}\) is (A) \(\mathrm{V}_{0} \cos \theta\) \((\mathrm{B})>\mathrm{V}_{0} \cos \theta\) (C) \(<\mathrm{V}_{0} \cos \theta\) (D) zero (ii) Maximum height h of the particle is $(\mathrm{A})=\left[\left(\mathrm{V}_{0}^{2} \sin ^{2} \theta\right) / 2 \mathrm{~g}\right]$ (B) $>\left[\left(\mathrm{V}_{0}^{2} \sin ^{2} \theta\right) / 2 \mathrm{~g}\right]$ $(\mathrm{C})<\left[\left(\mathrm{V}_{0}^{2} \sin ^{2} \theta\right) / 2 \mathrm{~g}\right]$ (D) can be greater than or less than $\left[\left(\mathrm{V}_{0}^{2} \sin ^{2} \theta\right) / 2 \mathrm{~g}\right]$

The distance of a geo-stationary satellite from the center of the earth (Radius \(\mathrm{R}=6400 \mathrm{~km}\) ) is nearest to (A) \(5 \mathrm{R}\) (B) \(7 \mathrm{R}\) (C) \(10 \mathrm{R}\) (D) \(18 \mathrm{R}\)

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