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Chapter 14: Problem 31

Show how, guided by Kepler's third law, Newton could deduce that the force holding the Moon in its orbit, assumed circular, must vary as the inverse square of the distance from the center of the Earth.

Short Answer

Expert verified
Newton deduced from Kepler's third law and the concept of centripetal force that the force holding the Moon in its orbit around the Earth, assumed circular, varies as the inverse square of the distance from the center of the Earth. That means gravitational force decreases with the square of the distance between the two objects.

Step by step solution

01

Understand Kepler's Third Law

Kepler's Third Law states that the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit. Mathematically, this can be written as: \(T^2 ∝ r^3\), where \(T\) is the orbital period and \(r\) is the distance from the center of the orbit.
02

Apply Kepler's Third Law to the Moon and Earth

Considering the Earth and the Moon, the relation between their orbital period \(T\) and the distance \(r\), according to Kepler's Third Law, is: \(T_{moon}^2 ∝ r_{moon}^3\). The Earth orbits the Sun in one year and the distance between them \(r_{earth}\), is one Astronomical Unit (AU), therefore: \(T_{earth}^2 ∝ r_{earth}^3\). We can now make the ratio between the Moon's and Earth's equation, which results in: \((\frac{T_{moon}}{T_{earth}})^2 ∝ (\frac{r_{moon}}{r_{earth}})^3\)
03

Use Centripetal Force Concept

Due to the circular movement of the Moon around the Earth, there is a centripetal force that keeps the Moon in its orbit and prevents it from going off straight. The centripetal force is given by: \(F = \frac{m*v^2}{r}\), where \(m\) is the mass of the object, \(v\) is the velocity and \(r\) is the radius of the motion. By expressing the velocity \(v\) in terms of the radius \(r\) and the period \(T\), we find: \(v = \frac{2*π*r}{T}\). Substituting this into the centripetal force equation gives: \(F ∝ \frac{1}{r^2}\)
04

Deduce the Inverse Square Law

By comparing the laws of the forces obtained from Kepler's Third Law and the concept of centripetal force, it can be noticed that the gravitational force must fall off as the square of the distance, known as the Inverse Square Law. This law states that the force of gravity acting between any two objects is inversely proportional to the square of the separation distance between the object's centers.

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Most popular questions from this chapter

Two point-like particles, each of mass \(m\), are originally separated by a distance \(d\) and moving in opposite directions each with a speed of \(v\). What is the maximum value for \(v\) so that the particles will eventually move back together under the influence of the mutual gravitational attraction?

Two neutron stars are separated by a center-to-center distance of \(93.4 \mathrm{~km}\). They each have a mass of \(1.56 \times 10^{30} \mathrm{~kg}\) and a radius of \(12.6 \mathrm{~km}\). They are initially at rest with respect to one another. (a) How fast are they moving when their separation has decreased to one-half of its initial value? \((b)\) How fast are they moving just before they collide? Ignore relativistic effects.

A comet moving in an orbit of eccentricity \(0.880\) has a speed of \(3.72 \mathrm{~km} / \mathrm{s}\) when it is most distant from the Sun. Find its speed when it is closest to the Sun.

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 \(K_{m} / K_{M}\).

The asteroid Eros, one of the many minor planets that orbit the Sun in the region between Mars and Jupiter, has a radius of \(7.0 \mathrm{~km}\) and a mass of \(5.0 \times 10^{15} \mathrm{~kg} .(a)\) If you were standing on Eros, could you lift a \(2000-\mathrm{kg}\) pickup truck? \((b)\) Could you run fast enough to put yourself into orbit? Ignore effects due to the rotation of the asteroid. (Note: The Olympic records for the \(400-\mathrm{m}\) run correspond to speeds of \(9.1 \mathrm{~m} / \mathrm{s}\) for men and \(8.2 \mathrm{~m} / \mathrm{s}\) for women.)
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