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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

A satellite revolves around the earth in an elliptical orbit. Its speed (A) is the same at all points in the orbit (B) is greatest when it is closest to the earth (C) is greatest when it is farthest to the earth (D) goes on increasing or decreasing continuously depending upon the mass of the satellite

The time period of a satellite of earth is 5 hours. If the separation between the earth and the satellite is increased to four times the previous value, the new time period will become \(\ldots \ldots \ldots\) hours (A) 10 (B) 120 (C) 40 (D) 80

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)

The time period \(\mathrm{T}\) of the moon of planet Mars \((\mathrm{Mm})\) is related to its orbital radius \(\mathrm{R}\) as \((\mathrm{G}=\) Gravitational constant \()\) (A) $\mathrm{T}^{2}=\left[\left(4 \pi^{2} \mathrm{R}^{3}\right) /(\mathrm{GMm})\right]$ (B) $\mathrm{T}^{2}=\left[\left(4 \pi^{2} \mathrm{GR}^{3}\right) /(\mathrm{Mm})\right]$ (C) \(T^{2}=\left[\left(2 \pi R^{2} G\right) /(M m)\right]\) (D) \(\mathrm{T}^{2}=4 \pi \mathrm{Mm} \mathrm{GR}^{2}\)

3 particle each of mass \(\mathrm{m}\) are kept at vertices of an equilateral triangle of side \(L\). The gravitational field at center due to these particles is (A) zero (B) \(\left[(3 \mathrm{GM}) / \mathrm{L}^{2}\right]\) (C) \(\left[(9 \mathrm{GM}) / \mathrm{L}^{2}\right]\) (D) \((12 / \sqrt{3})\left(\mathrm{Gm} / \mathrm{L}^{2}\right)\)
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