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Chapter 6: Problem 782

Potential energy of a satellite having mass \(\mathrm{m}\) and rotating at a height of \(6.4 \times 10^{6} \mathrm{~m}\) from the surface of earth (A) \(-0.5 \mathrm{mg} \operatorname{Re}\) (B) \(-\mathrm{mg} \mathrm{Re}\) (C) \(-2 \mathrm{mgRe}\) (D) \(4 \mathrm{mgRe}\)

Short Answer

Expert verified
First, calculate the satellite's distance from the center of the earth: \[r = \text{height} + \text{Re} = (6.4 \times 10^{6} ) + (6.371 \times 10^{6})\] Next, calculate the potential energy using the formula: \[U = -\frac{GMm}{r}\] Finally, compare the calculated potential energy with the given options to find the correct answer.

Step by step solution

01

Calculate the satellite's distance from the center of the earth

Since the satellite is at a height of \(6.4 \times 10^{6}\) m from the surface and Earth's radius is approximately \(6.371 \times 10^{6}\) m, the distance from the satellite to the center of the earth is: \[r = \text{height} + \text{Re} = (6.4 \times 10^{6} ) + (6.371 \times 10^{6})\]
02

Calculate the potential energy

Using the given mass of the satellite, and the calculated value of `r`, use the potential energy formula: \[U = -\frac{GMm}{r}\] Plug in the values for the gravitational constant \(G\), mass of the earth \(M\), and mass of the satellite \(m\).
03

Compare the result with the given options

Once the potential energy is calculated, compare the result with the four given options (A, B, C, and D), to find which option matches the calculated value of the potential energy.

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Most popular questions from this chapter

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) \(\mathrm{E}_{0}\)

If \(\mathrm{r}\) denotes the distance between the sun and the earth, then the angular momentum of the earth around the sun is proportional to (A) \(1^{3 / 2}\) (B) \(\mathrm{r}\) (C) \(r^{1 / 2}\) (D) \(r^{2}\)

If the earth is at one- fourth of its present distance from the sun the duration of year will be (A) half the present Year (B) one-eight the present year (C) one-fourth the present year (D) one-sixth the present year

When a particle is projected from the surface of earth, it mechanical energy and angular momentum about center of earth at all time is constant (i) A particle of mass \(\mathrm{m}\) is projected from the surface of earth with velocity \(\mathrm{V}_{0}\) at angle \(\theta\) with horizontal suppose \(\mathrm{h}\) be the maximum height of particle from surface of earth and \(\mathrm{v}\) its speed at that point them \(\mathrm{V}\) is (A) \(\mathrm{V}_{0} \cos \theta\) \((\mathrm{B})>\mathrm{V}_{0} \cos \theta\) (C) \(<\mathrm{V}_{0} \cos \theta\) (D) zero (ii) Maximum height h of the particle is $(\mathrm{A})=\left[\left(\mathrm{V}_{0}^{2} \sin ^{2} \theta\right) / 2 \mathrm{~g}\right]$ (B) $>\left[\left(\mathrm{V}_{0}^{2} \sin ^{2} \theta\right) / 2 \mathrm{~g}\right]$ $(\mathrm{C})<\left[\left(\mathrm{V}_{0}^{2} \sin ^{2} \theta\right) / 2 \mathrm{~g}\right]$ (D) can be greater than or less than $\left[\left(\mathrm{V}_{0}^{2} \sin ^{2} \theta\right) / 2 \mathrm{~g}\right]$

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