# Gravitation and Orbital Motion

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## Gravitation and Orbital Motion Answer Key

1.

Astronomers using Hubble Space Telescope discover a planet orbiting a distant star. They measure the planet's period of revolution to be equal to 135 days and the average distance between the planet and the star to be equal to [math]2.35xx10^8km[/math]. Calculate the planet's mass using this data.

- The force of gravity between the star and the planet must be equal to the centripetal force acting on the planet as it orbits the star. [math]G(M_sM_p)/r^2=(M_pv^2)/r[/math]. Mass of the planet [math]M_p[/math] cancels out. Solving for the mass of the star [math]M_s[/math] we obtain: [math]M_s=(v^2r)/G[/math]. The orbital speed [math]v[/math] can be determined by dividing the orbit's circumference by its period, [math]v=(2pir)/T[/math]. Substituting this expression into the equation for the star's mass we obtain:

[math]M_s=(4pi^2r^3)/(GT^2)[/math].

Substituting 135 days, or [math]1.167xx10^7 s[/math] for [math]T[/math], and [math]2.35xx10^11 m[/math] for [math]r[/math], and [math]6.67xx10^-11(m^3)/(kgs^2[/math] for [math]G[/math], we obtain [math]5.64xx10^31 kg[/math].

2.

Two identical asteroids of mass [math]M[/math] are viewed as spheres of radius [math]r[/math]. They are initially at rest at a distance orders of magnitude greater than their radii, and are being pulled towards each other by their own gravity. Assuming that no other objects are acting on them, what is the speed of the asteroids immediately prior to their collision? Assume this speed is much less than the speed of light.

- The kinetic energy of each of the asteroids is equal to the work done on it by the force of gravity:

[math](Mv^2)/2=int_(2r)^ooGM^2/r^2dr[/math]

Dividing both sides by the mass and multiplying by 2, we obtain:

[math]v^2=2GMint_(2r)^oo(dr)/r^2=(GM)/r[/math]

Therefore, [math]v=sqrt((GM)/r[/math].

3.

In order to get to the other side of the globe, an explorer digs a straight tunnel that goes through Earth's core and comes out on the other side. Describe the motion of an object dropped into this tunnel. Neglect air drag. If the object exhibits cyclical motion, calculate its period. Assume Earth's density to be homogeneous.

- The object will gain speed as the force of gravity does work on it. When the object reaches the Earth's core, the force of gravity becomes zero. The object will continue traveling through the tunnel, while being decelerated by the force of gravity. By the time the object reaches the other side, its speed will go back to zero, and the force of gravity will start pulling it into the tunnel. Therefore, the motion of the object will be cyclical.

Use Newton's Second law to mathematically describe the object's motion.

Let [math]M_E[/math] be the math of the Earth, [math]R_E[/math] - the Earth's radius, [math]m[/math] - the object's mass (it will cancel out in the end), and [math]r[/math] - the distance between the object and the Earth's core.

[math]m(d^2r)/(dt^2)=F_g[/math], where [math]F_g[/math] represents the force of gravity acting on the object.

According to the Newton's Law of Gravity, only the part of the Earth that is closer to the core than the object will exert any gravitational pull.

[math]F_g=-G(M_E(r^3/R_E^3)m)/r^2=-G(M_Emr)/R_E^3[/math]

Therefore, Newton's Second Law becomes [math](d^2r)/(dr^2)=-GM_E/R_E^3r[/math], which is a standard equation of the Simple Harmonic Motion with the period

[math]T=2pisqrt((R_E^3)/(GM_E)[/math] or 84 minutes and 18 seconds. Note that this is equal to the period of orbital motion at Earth's surface.

4.

One of the theories explaining the formation of Saturn rings shown below states that if an orbiting object gets close to a massive planet, the tidal force exerted on the object's surface will exceed the force of the object's own gravity, leading to the object's disintegration and eventual scattering along its orbit. The maximum distance from the planet for which this is possible is known as Roche limit after French astronomer Edouard Roche who first calculated this limit in 1848. Calculate Roche limit in a simplified case by equating the force of gravity exerted on a probe mass by a planet of mass [math]M_p[/math] and radius [math]R_p[/math] to the net force exerted on the same probe mass by a satellite of mass [math]M_s[/math] and radius [math]R_s[/math] orbiting the planet on a circular orbit of radius [math]d[/math] (Roche limit). Assume that the satellite is orbiting with synchronous rotation, meaning that it makes one rotation about its axis in about the same time it takes it to orbit the planet, much like the Moon orbiting Earth. Take the centrifugal force of this rotation into account.

- The gravitational pull [math]F_G[/math] on the probe mass [math]m[/math] towards the satellite is given by Newton's law of gravity: [math]F_G=G(M_sm)/R_S^2[/math]

This force has to be equal to [math]F_T+F_C[/math], the sum of the tidal force exerted by the planet and the centrifugal force due to the rotation of the satellite about its axis. The tidal force equals the difference in the planet's gravitational pull on the center of the satellite and on the edge of the satellite closest to the planet.

[math]F_T=G(M_p m)/((d-R_p)^2)-G(M_p m)/d^2[/math]

Reducing this expression to common denominator, we obtain:

[math]F_T=GM_p m(2dR_s-R_s^2)/(d^4-2d^3R_s+R_s^2d^2)[/math]

In the approximation where [math]R_s[/math] is much less than [math]R_p[/math] and [math]R_p[/math] is much less than [math]d[/math], we can say that the [math]R_p^2[/math] in the numerator and every term with [math]R_p[/math] in the denominator goes to zero, which gives us:

[math]F_T=(2GM_p mR_p)/d^3[/math]

For the centrifugal force from synchronous rotation we obtain:

[math]F_C=omega^2mR_p=GM_p mR_p/d^3[/math]

Equating [math]F_T+F_C[/math] with [math]F_G[/math] and solving for [math]d[/math] we obtain:

[math]d=R_p(3M_p/M_s)^(1/3)[/math]

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