Gravity of Extended Bodies
Problems
practice

Almost all stars will die as hot, dense lumps, but some of them decide to get fat and cold for awhile first. In 5 billion years our sun will expand into a red giant (red because it will be much cooler) swallowing up Mercury, Venus, Earth, and (possibly) Mars. A few hundred million years after that, it will begin shrinking until it becomes a white dwarf (white because it will be much, much hotter) about as small as the Earth but still retaining much of it mass.
Pairs of stars in orbit around one another are common in our galaxy — more common than lone stars like our sun. The stars never have the same mass and one will always age faster than the other. The diagram above shows a binary system where one star is a red giant and the other is a white dwarf. If the two stars are close enough, the white dwarf may begin "feeding" on the red giant. The matter torn off the surface of the red giant spirals into the white dwarf forming what is known as an accretion disk. Determine the separation between the two stars shown in this diagram. Let…
r = separation between the stars a = radius of red giant b = radius of white dwarf m_{a} = mass of red giant m_{b} = mass of white dwarf  Suppose that the Earth was an infinite flat slab of thickness t with the same mean density as the Earth. Calculate t in order that this infinite flat Earth has the same acceleration due to gravity on its surface as is found on the actual spherical Earth.
 Solve it once the easy way, using Gauss's law for gravity.
∯ g · dA = −4πGm
 Solve it once the hard way, using Newton's law of universal gravitation for continuous distributions of matter.
g(r) = − G ⌠⌠⌠
⌡⌡⌡r̂ dm r^{2}  How does this number compare to the radius of the actual spherical Earth?
 Solve it once the easy way, using Gauss's law for gravity.
 Dark matter
 The orbital speed of the planets decreases with distance from the Sun. Why does this happen? Derive a formula that shows the relationship.
 The orbital speed of the stars remains roughly constant with distance from the center of the Milky Way. (This is true for other galaxies as well.) What does this tell us about the distribution of mass in galaxies? Derive a formula that shows the relationship.
 Calculate the mass of the Milky Way given an orbital speed of 230 km/s and a radius of 50,000 light years. Give your answer in solar masses (m_{☉} = 1.99 × 10^{30} kg) and compare it to the number of stars in the Milky Way (~10^{11}).
 Dwarf galaxies, star clusters, and gas clouds beyond the edge of the visible galaxy have nearly the same orbital speed as the stars within visible galaxy. There is evidence that rotational speeds remain roughly constant at 220 km/s out to distances of 300,000 light years or six times the radius of the Milky Way. What is so amazing about this observation and what does it imply?
 The word nebula (plural nebulae) means cloud in latin. In astronomy, a nebula is a diffuse collection gas and dust that looks something like a cloud. Nebulae are larger than stars, but smaller than galaxies — on the order of 101000 solar systems in diameter. A few representative images are shown below.
A simplified model of a nebula is a spherical collection of matter whose density varies linearly from a maximum at its center to zero at its "surface". Determine the following quantities both inside and outside such a simplified nebula in terms of its radius R, the distance from the center r, the density at the center ρ_{0}, and fundamental constants…
 density
 gravitational field strength
 gravitational potential energy per unit mass
numerical
 Prolate spheroid
The image below shows a typical alignment of the three most popular objects in the solar system: the Sun, Earth, and Moon (not to scale). The letters around the Earth represent various locations on its surface.A = point closest to the Moon B = point farthest from the Moon C = point closest to the Sun D = point farthest from the Sun  Complete the following table.
 Use the results of your calculations to explain the effects that the Moon and Sun have on the shape of the Earth.
quantity Moon Sun mass (kg) 7.348 × 10^{22} 1.989 × 10^{30} distance from Earth (m) 3.844 × 10^{8} 1.496 × 10^{11} gravitational field (N/kg) from object on Earth, at Earth's center (g) on Earth, nearest to object (g_{near}) on Earth, farthest from object (g_{far}) ratio of difference in field at opposite
sides of the Earth to field at center (1:x)  Oblate spheroid
Compare these images of Earth and Jupiter (not to scale). Complete the following table.
 Use the results of your calculations to explain the effect that rotation has on the shapes of Earth and Jupiter.
quantity Earth Jupiter mass (kg) 5.9742 × 10^{24} 1.8988 × 10^{27} period of rotation (h) 23.935 9.9250 radius (m) polar 6,356,750 66,854,000 equatorial 6,378,140 71,492,000 oblateness: ratio of difference
in radii to equatorial radius (1:x)rotational speed (m/s) at either pole on the equator centrifugal acceleration (m/s^{2}) at either pole on the equator gravitational acceleration (m/s^{2}) at either pole on the equator ratio of centrifugal to gravitational
accelerations at the equator (1:x)
investigative
 Determine the apparent acceleration due to gravity at your current location on Earth. Here is the general procedure for doing this.
 Start by determining your latitude and altitude. That information is probably available somewhere on the internet or you could measure it yourself with a GPS enabled device.
 The Earth is an oblate spheroid. Use the following fancy formula to calculate your distance (r) from the center of the Earth…
r = h + √ (a^{2} cos φ)^{2} + (b^{2} sin φ)^{2} (a cos φ)^{2} + (b sin φ)^{2} h = your altitude φ = your latitude a = earth's equatorial radius (6,378,137 m) b = earth's polar radius (6,356,752 m)  Compute the magnitude of the gravitational acceleration at your location.
 Compute your distance from the Earth's axis; that is, the component of r parallel to the plane of the equator.
 Compute the magnitude of the centrifugal acceleration at your location.
 Combine the gravitational and centrifugal components to determine the apparent acceleration due to gravity (g′) at your location. Determine both the…
 magnitude and
 direction (relative to r, the vector that points directly toward the center of the Earth)
 Compare your results to the values often found in textbooks and reference tables.
 9.8 m/s^{2} (this value with its two significant digits of accuracy should agree with your results)
 9.798 m/s^{2} (an average value calculated over the surface of the Earth)
 9.80665 m/s^{2} (the defined unit known by the short name of standard gravity)
algebraic
 Determine the rotation curve for…
 a spherical galaxy with a uniform mass distribution
 a thin, diskshaped galaxy with a uniform mass distribution
calculus
 Given the nebula in practice problem 4, determine the…
 location
 value
 Determine the gravitational field and gravitational potential, inside and outside the following mass distributions…
 a sphere of mass M and uniform density ρ.
 a simplified model of the Earth, whose total mass M is evenly split between the core and the mantle — ½M for each part. (Assume the crust, oceans, and atmosphere make a negligible contribution to Earth's mass.)
 a simplified model of the Earth consisting of a core with density 2ρ and mantle with density ρ. (Assume the crust, oceans, and atmosphere make a negligible contribution to the Earth's mass.)
 a galaxy (including dark matter halo) with a density that decreases according to the function ρ = ρ_{0}/r^{2}, where r is the distance from the galactic center.