**Unformatted text preview: **reefall (page 379). Thus the statement is incorrect.
b. They are not beyond the pull of the Earth’s gravity, as plugging in numbers into Newton’s Law of
Universal Gravitation will give you non-zero answers. What is happening instead is that they are
constantly falling, but in such a way they are orbiting the planet and not rapidly approaching the ground. Problem 7)
We can set up the equation in the following manner: Like problem 4, we have to set up this equation in such a manner that we can use the Pythagorean
theorem to solve for h. By the Pythagorean Theorem: Problem 8)
a. We combine Newton’s First Law and the Law of Universal Gravitation: b. Here, we are basically asked to use Kepler’s Third Law (12-7) to calculate the period such that loose
rocks on do not fly off the surface. We divide the answer by 3600 (because the answer will be in seconds, and we want the units in hours),
which gives us 9.3 hours. Problem 9)
a. According to Kepler’s Second Law, as a planet moves in its orbit, it sweeps out an equal amount of
area in an equal amount of time. For this to be true, that means the planet must move the slower when
it is furthest away from the sun and fastest when it close to the sun. Thus, on January 4th the distance of
the Earth to the Sun is less than the distance of the Earth to the Sun on July 4th, as the orbital speed is
greatest around January 4th.
b. See above. Problem 10)
a. If we ascribe to a lunar month (which is based on the complete orbit of the moon around the earth),
increasing distance between the earth and moon implies a increase in the lunar month by Kepler’s Third
Law. b. The greater the radius of an orbit, the greater the period, which implies a longer month. Problem 11)
We use Kepler’s Third Law to solve for the Mass of Jupiter: Problem 12)
Let’s consider: If an orbit of a geosynchronous satellite is 24 hours,...

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