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Chapter 14: Problem 24

Determine the mass of the Earth from the period \(T\) and the radius \(r\) of the Moon's orbit about the Earth: \(T=27.3\) days and \(r=3.82 \times 10^{5} \mathrm{~km}\).

Short Answer

Expert verified
The calculated mass of the Earth, using the data provided for the period and radius of the Moon's orbit, is approximately \(5.98 \times 10^{24}\) kg.

Step by step solution

01

Understand the problem

Given the period, \(T\), and the radius, \(r\), of the Moon's orbit, the task is to find the mass of the Earth. From Kepler's third law, we know \(T^2 \propto r^3\). Since the proportionality constant is the same for all bodies in the universe, we will equate this to the gravitational force exerted by the Earth on the Moon.
02

Convert units

Firstly, it's important to convert the given units into SI units. The period \(T\) is given in days. So convert it into seconds: 1 day = 86400 seconds. So \(T = 27.3 \times 86400 \) s. The distance \(r\) is given in kilometers, so it needs to be converted to meters: \(r = 3.82 \times 10^{5} \times 10^{3}\) m.
03

Apply Kepler's third law

According to Kepler's Third law, \(((2\pi)^2/GM)R^3 = T^2\). Solve for \(M\) (the mass of the Earth) to get \(M= \((2\pi)^2/G)R^3/ T^2\). Here, \(G\) is the gravitational constant, equal to \(6.67 \times 10^{-11} m^3 kg^{-1} s^{-2}\). \(R\), as mentioned earlier, is the radius of orbit, and \(T\) is the period of orbit.
04

Substituting the known values into the equation

We substitute the known values into the equation for mass derived in the previous step to find the mass of the Earth: \(M= \((2\pi)^2/G)R^3/ T^2\).
05

Perform Calculations

After substituting the relevant values for \(T\), \(R\), and \(G\) the calculations yield the resultant mass of Earth.

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Most popular questions from this chapter

Show that the velocity of escape from the Sun at the Earth's distance from the Sun is \(\sqrt{2}\) times the speed of the Earth in its orbit, assumed to be a circle. (This is a specific case of a general result for circular orbits: \(v_{\text {esc }}=\sqrt{2} v_{\text {orb }} .\) )

A pair of stars revolves about their common center of mass, as in Fig. \(14-31\). One of the stars has a mass \(M\) that is twice the mass \(m\) of the other; that is, \(M=2 m .\) Their centers are a distance \(d\) apart, \(d\) being large compared to the size of either star. (a) Derive an expression for the period of revolution of the stars about their common center of mass in terms of \(d, m\), and \(G .(b)\) Compare the angular momenta of the two stars about their common center of mass by calculating the ratio \(L_{m} / L_{M} .(c)\) Compare the kinetic energies of the two stars by calculating the ratio \(K_{m} / K_{M}\).

(a) Does it take more energy to get a satellite up to \(1600 \mathrm{~km}\) above the Earth than to put it in orbit once it is there? (b) What about \(3200 \mathrm{~km} ?(c)\) What about \(4800 \mathrm{~km}\) ? Take the Earth's radius to be \(6400 \mathrm{~km}\).

Express the universal gravitational constant \(G\) that appears in Newton's law of gravity in terms of the astronomical unit \(\mathrm{AU}\) as a length unit, the solar mass \(M_{\mathrm{S}}\) as a mass unit, and the year as a time unit. \(\left(1 \mathrm{AU}=1.496 \times 10^{11} \mathrm{~m}, 1 \mathrm{M}_{\mathrm{S}}=1.99 \times\right.\) \(\left.10^{30} \mathrm{~kg}, 1 \mathrm{y}=3.156 \times 10^{7} \mathrm{~s} .\right)(b)\) What form does Kepler's third law (Eq. \(14-23\) ) take in these units?

If \(g\) is to be determined by dropping an object through a distance of (exactly) \(10 \mathrm{~m}\), how accurately must the time be measured to obtain a result good to \(0.1 \%\) ? Calculate a percent error and an absolute error, in milliseconds.
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