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Chapter 6: Problem 781

A satellite with K.E. \(E_{\mathrm{k}}\) is revolving round the earth in a circular orbit. How much more K.E. should be given to it so that it may just escape into outer space ? (A) \(\mathrm{E}_{\mathrm{k}}\) (B) \(2 \mathrm{E}_{\mathrm{k}}\) (C) \((1 / 2) \mathrm{E}_{\mathrm{k}}\) (D) \(3 \mathrm{E}_{\mathrm{k}}\)

Short Answer

Expert verified
The additional kinetic energy required for the satellite to escape Earth's gravity is equal to its current kinetic energy (\(E_{k}\)). Therefore, the correct answer is (A) \(E_{k}\).

Step by step solution

01

Find escape velocity formula

The formula to calculate escape velocity is given by: \(v_{e} = \sqrt{\frac{2GM}{R}}\), where \(v_{e}\) is the escape velocity, \(G\) is the gravitational constant, \(M\) is the mass of the Earth, and \(R\) is the distance from the center of the Earth to the satellite, which is the sum of Earth's radius and the altitude of the satellite.
02

Find orbital velocity formula

The formula to calculate orbital velocity is given by: \(v_{o} = \sqrt{\frac{GM}{R}}\), where \(v_{o}\) is the orbital velocity.
03

Calculate kinetic energies using velocities

The kinetic energy of the satellite in its current state is given by \(E_{k} = \frac{1}{2}mv_{o}^2\), and the kinetic energy it needs to have to escape Earth's gravity is given by \(E_{k_{e}} = \frac{1}{2}mv_{e}^2\).
04

Substitute the expressions for velocities

Now, substitute the expressions for \(v_{o}\) and \(v_{e}\) from steps 1 and 2 into the kinetic energy equations: \(E_{k} = \frac{1}{2}m\left(\sqrt{\frac{GM}{R}}\right)^2\) \(E_{k_{e}} = \frac{1}{2}m\left(\sqrt{\frac{2GM}{R}}\right)^2\)
05

Simplify the expressions for kinetic energies

Upon simplification, we get: \(E_{k} = \frac{1}{2}m\left(\frac{GM}{R}\right)\) \(E_{k_{e}} = \frac{1}{2}m\left(\frac{2GM}{R}\right)\)
06

Find the additional kinetic energy required

The additional kinetic energy required is the difference between \(E_{k_{e}}\) and \(E_{k}\): Additional Kinetic Energy = \(E_{k_{e}} - E_{k}\) = \(\frac{1}{2}m\left(\frac{2GM}{R}\right) - \frac{1}{2}m\left(\frac{GM}{R}\right)\) By simplifying the equation, we get: Additional Kinetic Energy = \(E_{k}\) So, the correct answer is (A) \(E_{k}\).

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Most popular questions from this chapter

The radius of orbit of a planet is two times that of earth. The time period of planet is \(\ldots \ldots \ldots\) years. (A) \(4.2\) (B) \(2.8\) (C) \(5.6\) (D) \(8.4\)

The acceleration of a body due to the attraction of the earth (radius R) at a distance \(2 \mathrm{R}\) from the surface of the earth is \(=\) (g \(=\overline{\text { acceleration due to gravity at the surface of earth })}\) (A) \(\mathrm{g} / 9\) (B) \(\mathrm{g} / 3\) (C) \(\mathrm{g} / 4\) (D) 9

Suppose the gravitational force varies inversely as the nth power of distance the time period of planet in circular orbit of radius \(\mathrm{R}\) around the sun will be proportional to (A) \(\mathrm{R}^{[(\mathrm{n}+1) / 2]}\) (B) \(\mathrm{R}^{[(\mathrm{n}-1) / 2]}\) (C) \(\mathrm{R}^{\mathrm{n}}\) (D) \(\mathrm{R}^{[(\mathrm{n}-1) / 2]}\)

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.

The period of a satellite in circular orbit around a planet is independent of (A) the mass of the planet (B) the radius of the planet (C) mass of the satellite (D) all the three parameters (A), (B) and (C)
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