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Chapter 6: Problem 734

If \(\mathrm{g}\) is the acceleration due to gravity at the earth's surface and \(\mathrm{r}\) is the radius of the earth, the escape velocity for the body to escape out of earth's gravitational field is \(\ldots \ldots \ldots\) (A) \(\mathrm{gr}\) (B) \(\sqrt{(2 \mathrm{gr})}\) (C) \(\mathrm{g} / \mathrm{r}\) (D) \(\mathrm{r} / \mathrm{g}\)

Short Answer

Expert verified
The short answer to the question is: The escape velocity for a body to escape the Earth's gravitational field is \(v_e = \sqrt{2gr}\), which corresponds to option (B).

Step by step solution

01

Understand the escape velocity formula

To find the escape velocity, we must understand its formula and the concepts behind it. Escape velocity is defined as the minimum velocity that an object must achieve to escape the gravitational pull of a celestial body. The formula for escape velocity is given by: \(v_e = \sqrt{\frac{2GM}{r}}\), where - \(v_e\) represents the escape velocity, - G is the gravitational constant, with a value of approximately \(6.674\times 10^{-11} \frac{m^3}{kg⋅s^2} \), - M is the mass of the celestial body (in our case, the Earth), - r is the distance from the center of the celestial body to the object (which is equal to the radius of the Earth for this problem).
02

Express g in terms of G and M

To proceed with the escape velocity formula in the given terms, we need to express gravitational acceleration (g) in terms of the gravitational constant (G) and the Earth's mass (M). The formula linking g, G, and M is given by: \(g = \frac{GM}{r^2}\).
03

Solve for escape velocity in terms of g and r

We will now substitute the expression for g from the above equation into the escape velocity formula. We have: \(v_e = \sqrt{\frac{2GM}{r}}\) By substituting \(g = \frac{GM}{r^2}\) into the escape velocity formula, we get: \(v_e = \sqrt{\frac{2g{r^2}}{r}}\)
04

Simplify the expression

By simplifying the escape velocity expression obtained above, we can get it in terms of g and r only: \(v_e = \sqrt{\frac{2g{r^2}}{r}} = \sqrt{2gr}\)
05

Choose the correct option

Comparing the simplified expression to the given options, we can find the correct answer. The escape velocity for a body to escape the Earth's gravitational field is: \(v_e = \sqrt{2gr}\) This matches option (B). So, the correct answer is: (B) \(\sqrt{2gr}\)

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Most popular questions from this chapter

A body weight \(500 \mathrm{~N}\) on the surface of the earth. How much would it weight half way below the surface of earth (A) \(125 \mathrm{~N}\) (B) \(250 \mathrm{~N}\) (C) \(500 \mathrm{~N}\) (D) \(1000 \mathrm{~N}\)

Given mass of the moon is \((1718)\) of the mass of the earth and corresponding radius is \((1 / 4)\) of the earth, If escape velocity on the earth surface is \(11.2 \mathrm{kms}^{-1}\) the value of same on the surface of moon is $=\ldots \ldots \ldots \ldots \mathrm{kms}^{-1}$. (A) \(0.14\) (B) \(0.5\) (C) \(2.5\) (D) 5

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The mass of a space ship is \(1000 \mathrm{~kg} .\) It is to be launched from earth's surface out into free space the value of \(\mathrm{g}\) and \(\mathrm{R}\) (radius of earth) are \(10 \mathrm{~ms}^{-2}\) and \(6400 \mathrm{~km}\) respectively the required energy for this work will be $=\ldots \ldots \ldots .$ J (A) \(6.4 \times 10^{11}\) (B) \(6.4 \times 10^{8}\) (C) \(6.4 \times 10^{9}\) (D) \(6.4 \times 10^{10}\)

If the height of a satellite from the earth is negligible in comparison of the radius of the earth \(\mathrm{R}\), the orbital velocity of the satellite is (A) \(\mathrm{gR}\) (B) \((\mathrm{gR} / 2)\) (C) \(\sqrt{(g} / \mathrm{R})\) (D) \(\sqrt{(g R)}\)
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