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Chapter 6: Problem 724

Escape velocity of a body of \(1 \mathrm{~kg}\) on a planet is $100 \mathrm{~ms}^{-1}$. Gravitational potential energy of the body at the planet is \(=\) $\begin{array}{ll}\text { (A) } \overline{-5000} & \text { (B) }-1000\end{array}$ (C) \(-2400\) (D) 5000

Short Answer

Expert verified
The gravitational potential energy of the body at the planet's surface is \(-5000\mathrm{~J}\). So, the correct answer is (A) \(-5000\).

Step by step solution

01

Formula for escape velocity

To find the gravitational potential energy, we will first look at the formula for escape velocity. The 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 planet, and \(r\) is the distance from the center of the planet to the body. Using the given escape velocity, \(100\mathrm{~ms^{-1}}\), we can calculate the value of \(\frac{2GM}{r}\).
02

Kinetic energy and gravitational potential energy

Next, we can relate kinetic energy (KE) and gravitational potential energy (GPE) to escape velocity. For a body to escape a planet's gravitational field, its kinetic energy needs to be equal to the gravitational potential energy: \(KE = GPE\) The kinetic energy of the body is given by: \(KE = \frac{1}{2}mv^2\) Using the escape velocity, the kinetic energy of the body is: \(KE = \frac{1}{2}(1\textrm{ kg})(100\textrm{ ms}^{-1})^2\) Now, the gravitational potential energy is: \(GPE = -\frac{GMm}{r}\) Since \(KE = GPE\), we have: \(\frac{1}{2}mv^2 = -\frac{GMm}{r} \Rightarrow \frac{1}{2}v^2 = -\frac{GM}{r}\)
03

Calculate the gravitational potential energy

Now we can solve for \(GPE\): \(-\frac{GM}{r} = \frac{1}{2}(100\textrm{ ms}^{-1})^2\) Let's plug in the value of \(\frac{2GM}{r}\) we got in step 1: \(GPE = -\frac{1}{2}(100\textrm{ ms}^{-1})^2 = -5000\mathrm{~J}\) The gravitational potential energy of the body at the planet's surface is -5000 J. So, the correct answer is (A) \(-5000\).

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Most popular questions from this chapter

A body is projected vertically upwards from the surtace of a planet of radius \(\mathrm{R}\) with a velocity equal to half the escape velocity for that planet. The maximum height attained by the body is $\ldots \ldots \ldots \ldots$ (A) \((\mathrm{R} / 3)\) (B) \((\mathrm{R} / 2)\) (C) \((\mathrm{R} / 4)\) (D) \((\mathrm{R} / 5)\)

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) \(\sqrt{(g R)}\)

A particle of mass \(\mathrm{M}\) is situated at the center of a spherical shell of same mass and radius a the magnitude of gravitational potential at a point situated at (a/2) distance from the center will be (A) \([(4 \mathrm{GM}) / \mathrm{a}]\) (B) \((\mathrm{GM} / \mathrm{a})\) (C) \([(2 \mathrm{GM}) / \mathrm{a}]\) (D) \([(3 \mathrm{GM}) / \mathrm{a}]\)

Direction (Read the following questions and choose) (A) If both Assertion and Reason are true and the Reason is correct explanation of assertion (B) If both Assertion and Reason are true, but reason is not correct explanation of the Assertion (C) If Assertion is true, but the Reason is false (D) If Assertion is false, but the Reason is true Assertion : Unit of gravitational field intensity is $\mathrm{N} / \mathrm{kg}\( or \)\mathrm{ms}^{-2}\( Reason: Gravitational field intensity \)[$ Force \() /(\) mass $)]=(\mathrm{N} / \mathrm{kg})=\left[\left(\mathrm{kg} \cdot \mathrm{m} / \mathrm{sec}^{2}\right) / \mathrm{kg}\right]=\mathrm{ms}^{-2}$ (a) \(\mathrm{A}\) (b) \(\mathrm{B}\) (c) \(\mathrm{C}\) (d) D

A small satellite is revolving near earth's surface. Its orbital velocity will be nearly \(=\ldots \ldots \ldots \ldots \mathrm{kms}^{-1}\). (A) 8 (B) 4 (C) 6 (D) \(11.2\)
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