Understanding Our Universe
Understanding Our Universe
3rd Edition
ISBN: 9780393614428
Author: PALEN, Stacy, Kay, Laura, Blumenthal, George (george Ray)
Publisher: W.w. Norton & Company,
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Chapter 3, Problem 25QAP
To determine

The justification for time period of Neptune around the sun in spite of larger orbital distance.

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Neptune orbits the Sun with an orbital radius of 4.495 x 10^12 m. If the earth to sun distance 1A.U. = 1.5 x 10^11 m, a) Determine how many A.U.'s is Neptune's orbital radius (Round to the nearest tenth). b) Given the Sun's mass is 1.99 x10^30 kg, use Newton's modified version of Kepler's formula T^2 = (4pi^2/Gm(star)) x d^3 to find the period in seconds using scientific notation. (Round to the nearest thousandth). C) Convert the period in part b) to years (Round to the nearest tenth)
A synchronous satellite, which always remains above the same point on a planet's equator, is put in circular orbit around Neptune so that scientists can study a surface feature. Neptune rotates once every 16.1 h. Use the data of this table to find the altitude of the satellite. km
While looking through the Mt. Palomar telescope, you discover a large planetary object orbited by a single moon. The moon orbits the planet every 7.35 hours with the centers of the two objects separated by a distance roughly 2.25 times the radius of the planet. Fellow scientists speculate that the planet is made of mostly iron. In fact, the media has dubbed it the ''Iron Planet'' and NASA has even named it Planet Hephaestus after the Greek god of iron. But you have your doubts. Assuming the planet is spherical and the orbit circular, calculate the density of Planet Hephaestus.
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