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Chapter 6: Problem 777

The distance of a geo-stationary satellite from the center of the earth (Radius \(\mathrm{R}=6400 \mathrm{~km}\) ) is nearest to (A) \(5 \mathrm{R}\) (B) \(7 \mathrm{R}\) (C) \(10 \mathrm{R}\) (D) \(18 \mathrm{R}\)

Short Answer

Expert verified
The distance of a geo-stationary satellite from the center of the Earth is approximately \(6.6 R_E\), where \(R_E\) represents Earth's radius (\(6400\,km\)). Comparing this result with the given options, the nearest option is (B) \(7 R\).

Step by step solution

01

Find the gravitational force acting on the satellite

The gravitational force acting on a satellite can be represented by the formula: \[F = G\frac{m_s m_E}{R^2}\] Where \(F\) is the gravitational force, \(G\) is the gravitational constant \((6.67\times10^{-11}N\cdot m^2/kg^2)\), \(m_s\) is the mass of the satellite, \(m_E\) is the mass of the Earth \((5.97\times10^{24}kg)\), and \(R\) is the distance between the centers of the Earth and the satellite.
02

Find the centripetal force on the satellite

The centripetal force keeping the satellite in orbit can be represented by the formula: \[F_c = m_s \frac{v^2}{R}\] Where \(F_c\) is the centripetal force, \(m_s\) is the mass of the satellite, \(v\) is the satellite's orbital velocity, and \(R\) is the distance between the centers of the Earth and the satellite. Since the satellite's orbital period is equal to the Earth's rotational period (24 hours), we can find its orbital speed using the formula: \[v = \frac{2\pi R}{T}\] Here, \(T\) is the orbital period which is equal to \(24 \times 3600\,s\).
03

Equate gravitational force and centripetal force

We can set the gravitational force equal to the centripetal force and solve for R. \[\frac{Gm_s m_E}{R^2} = m_s \frac{v^2}{R}\] Since the mass of the satellite, \(m_s\), appears on both sides of the equation, we can cancel it out: \[\frac{G m_E}{R^2} = \frac{v^2}{R}\]
04

Solve for R

Substitute the formula for orbital velocity, \(v\), in the equation and solve for R. \[\frac{G m_E}{R^2} = \frac{(2\pi R/T)^2}{R}\] \[\frac{G m_E}{R^2} = \frac{4\pi^2 R^2}{T^2}\] Now, multiply both sides by R^3 and rearrange the equation: \[R^3 = \frac{G m_E T^2}{4\pi^2}\] Plug in the values of G, m_E, and T, and solve for R: \[R^3 = \frac{(6.67\times10^{-11})(5.97\times10^{24})(24\times3600)^2}{4\pi^2}\] \[R \approx 4.22\times10^7 m\] Now, we need to compare this with Earth's radius (\(R_E = 6400\,km = 6.4\times10^6 m\)): \[\frac{R}{R_E} \approx \frac{4.22\times10^7}{6.4\times10^6}\approx 6.6\]
05

Choose the appropriate option

Based on our calculation, the distance of a geo-stationary satellite from the center of the Earth is closest to \(6.6 R_E\). Comparing this result with the given options, the nearest option is (B) \(7 R\).

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