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Chapter 6: Problem 739

The escape velocity of a body from earth's surface is Ve. The escape velocity of the same body from a height equal to 7 R from earth's surface will be (A) \((\mathrm{Ve} / \sqrt{2})\) (B) \((\mathrm{Ve} / 2)\) (C) \((\mathrm{Ve} / 2 \sqrt{2})\) (D) \((\mathrm{Ve} / 4)\)

Short Answer

Expert verified
The escape velocity of the same body from a height equal to 7R from Earth's surface is \(V_{e1} = \frac{V_e}{2}\). The correct answer is option (B).

Step by step solution

01

Write the formula for escape velocity from Earth's surface.

Using the formula for escape velocity: \(V_e = \sqrt{\frac{2GM}{r_e}}\) Here, the mass of Earth is M and the distance from the Earth's center to its surface (Earth's radius) is R.
02

Write the formula for escape velocity at a height of 7R from Earth's surface.

Now, we will write the formula for escape velocity at a height of 7R from the Earth's surface. Let the new escape velocity be \(V_{e1}\). The distance from the Earth's center at this height becomes r = 8R. \(V_{e1} = \sqrt{\frac{2GM}{8R}}\)
03

Compare the two escape velocities.

Now we need to find the relation between \(V_e\) and \(V_{e1}\). To do this, we will divide the equations of \(V_{e1}\) by \(V_e\). \(\frac{V_{e1}}{V_e} = \frac{\sqrt{\frac{2GM}{8R}}}{\sqrt{\frac{2GM}{R}}}\)
04

Simplify the equation to find the ratio between the two escape velocities.

Simplify the equation to find the ratio between \(V_{e1}\) and \(V_e\): \(\frac{V_{e1}}{V_e} = \frac{\sqrt{\frac{1}{4}}}{\sqrt{1}} = \frac{1}{2}\) So, \(V_{e1} = \frac{V_e}{2}\). Therefore, the escape velocity of the same body from a height equal to 7R from Earth's surface is: \(V_{e1} = \frac{V_e}{2}\) The correct answer is option (B).

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Most popular questions from this chapter

A body weights \(700 \mathrm{~g} \mathrm{wt}\) on the surface of earth. How much it weight on the surface of planet whose mass is \(1 / 7\) and radius is half that of the earth (A) \(200 \mathrm{~g} \mathrm{wt}\) (B) \(400 \mathrm{~g} \mathrm{wt}\) (C) \(50 \mathrm{~g} \mathrm{wt}\) (D) \(300 \mathrm{~g}\) wt.

An artificial satellite moving in a circular orbit around earth has a total (kinetic + potential energy) \(E_{0}\), its potential energy is (A) \(-\mathrm{E}_{0}\) (B) \(1.5 \mathrm{E}_{0}\) (C) \(2 \mathrm{E}_{0}\) (D) \(\mathrm{E}_{0}\)

Match the following Table-1 \(\quad\) Table-2 (A) kinetic energy (P) \([(-\mathrm{GMm}) /(2 \mathrm{r})]\) (B) Potential energy (Q) \(\sqrt{(\mathrm{GM} / \mathrm{r})}\) (C) Total energy (R) - [(GMm) / r] (D) orbital velocity (S) \([(\mathrm{GMm}) /(2 \mathrm{r})]\) Copyright (O StemEZ.com. All rights reserved.

If the radius of the earth were to shrink by \(1 \%\) its mass remaining the same, the acceleration due to gravity on the earth's surface would (A) decrease by \(2 \%\) (B) remain Unchanged (C) increase by \(2 \%\) (D) increases by \(1 \%\)

As astronaut orbiting the earth in a circular orbit \(120 \mathrm{~km}\) above the surface of earth, gently drops a spoon out of space-ship. The spoon will (A) Fall vertically down to the earth (B) move towards the moon (C) Will move along with space-ship (D) Will move in an irregular wav then fall down to earth
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