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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

The escape velocity for a body projected vertically upwards from the surface of earth is \(11 \mathrm{kms}^{-1}\). If the body is projected at an angle of \(45^{\circ}\) with the vertical, the escape velocity will be $\ldots \ldots \ldots \mathrm{kms}^{-1}$ (A) \((11 / \sqrt{2})\) (B) \(11 \sqrt{2}\) (C) 22 (D) 11

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) \(6.4 \times 10^{10}\)

There are two planets, the ratio of radius of two planets is \(\mathrm{k}\) but the acceleration due to gravity of both planets are \(\mathrm{g}\) what will be the ratio of their escape velocity. (A) \((\mathrm{kg})^{1 / 2}\) (B) \((\mathrm{kg})^{-1 / 2}\) (C) \((\mathrm{kg})^{2}\) (D) \((\mathrm{kg})^{-2}\)

According to keplar, the period of revolution of a planet ( \(\mathrm{T}\) ) and its mean distance from the sun (r) are related by the equation (A) \(\mathrm{T}^{3} \mathrm{r}^{3}=\) constant (B) \(\mathrm{T}^{2} \mathrm{r}^{-3}=\) constant (C) \(\mathrm{Tr}^{3}=\) constant (D) \(\mathrm{T}^{2} \mathrm{r}=\) constant

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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