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Chapter 6: Problem 801

A satellite of mass \(\mathrm{m}\) is circulating around the earth with constant angular velocity. If radius of the orbit is \(\mathrm{R}_{0}\) and mass of earth \(\mathrm{M}\), the angular momentum about the center of earth is (A) \(m \sqrt{\left(G M R_{0}\right)}\) (B) \(\mathrm{M} \sqrt{\left(\mathrm{GMR}_{\mathrm{o}}\right)}\) (C) \(\left.m \sqrt{(G M} / R_{0}\right)\) (D) \(\mathrm{M} \sqrt{\left(\mathrm{GM} / \mathrm{R}_{\mathrm{o}}\right)}\)

Short Answer

Expert verified
The short answer is: The angular momentum of the satellite about the center of Earth is given by \(L = m\sqrt{GMR_0}\), which corresponds to option (A).

Step by step solution

01

Find the linear velocity of the satellite

The linear velocity, \(v\), of an object in circular motion can be given by: \(v = ωr\) where \(ω\) is the angular velocity, and \(r\) is the radius of the circle (in this case, the orbit of the satellite).
02

Relate linear velocity to gravitational force

The centripetal force acting on the satellite is equal to the gravitational force acting on it. We can use Newton's law of gravitation to write this force as: \(F = \frac{GmM}{r^2}\) where \(G\) is the gravitational constant, \(m\) is the mass of the satellite, \(M\) is the mass of the Earth, and \(r\) is the radius of the circular orbit (\(R_0\)). In circular motion, centripetal force can also be written as: \(F = \frac{mv^2}{r}\) Equating these two expressions for the centripetal force gives: \(\frac{GmM}{r^2} = \frac{mv^2}{r}\)
03

Solve for linear velocity

We can now solve the equation for the linear velocity, \(v\), as follows: \(v^2 = \frac{GmMr}{r^3}\) \(v^2 = \frac{GM}{r}\) \(v = \sqrt{\frac{GM}{r}}\)
04

Compute the angular momentum

Now that we have the expression for the linear velocity, we can substitute it back into the formula for the angular momentum: \(L = mvr\) \(L = m\cdot\sqrt{\frac{GM}{r}}\cdot r\) Since \(r = R_0\), \(L = m\sqrt{GM R_0}\) Comparing our result with the given answer options, we find that the correct expression for the angular momentum of the satellite is \(L = m\sqrt{GMR_0}\), which corresponds to option (A).

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Most popular questions from this chapter

The mass and radius of the sun are \(1.99 \times 10^{30} \mathrm{~kg}\) and \(\mathrm{R}=6.96 \times 10^{8} \mathrm{~m}\). The escape velocity of rocket from the sun \(\mathrm{is}=\ldots \ldots \ldots \mathrm{km} / \mathrm{sec}\) $\begin{array}{llll}\text { (A } 11.2 & \text { (B) } 12.38 & \text { (C) } 59.5 & \text { (D) } 618\end{array}$

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The radii of two planets are respectively \(\mathrm{R}_{1}\) and \(\mathrm{R}_{2}\) and their densities are respectively \(\rho_{1}\) and \(\rho_{2}\) the ratio of the accelerations due to gravity at their surface is (A) $g_{1}: g_{2}=\left(\rho_{1} / R_{1}^{2}\right) \cdot\left(\rho_{2} / R_{2}^{2}\right)$ (B) $\mathrm{g}_{1}: \mathrm{g}_{2}=\mathrm{R}_{1} \mathrm{R}_{2}: \rho_{1} \rho_{2}$ (C) \(g_{1}: g_{2}=R_{1} \rho_{2} \cdot R_{2} p_{1}\) (D) \(g_{1}: g_{2}=R_{1} \rho_{1}: R_{2} \rho_{2}\)
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