3. The Moon has a period of 27.3 days and a mean distance of 3.90×105 km fromthe center of Earth.a. Use Kepler's laws to find the period of a satellite in orbit 6.70x103 kmfrom the center of Earth.b. How far above Earth's surface is this satellite? Law of Harmonies:The periods of the planets are proportional to the 3/2 powers ofthe major axis lengths of their orbits.To explain further, the law states that the square of the time it takes for the planetto make one revolution around the sun (known as the orbital period) isproportional to the cube of the semimajor axis of the planet's elliptical orbit.where,T= orbital periods of a planet (planet 1 or 2)a = length of the semimajor axisUniversal Gravitation and Kepler's Third LawWe can associate Newton's law of universal gravitation to Kepler's law of harmoniessince we can apply these equations to the motion of planets about the Sun.Suppose a planet is orbiting the Sun. Using Newton's second law of motion, we have:Fnet = maWe can rewrite the equation in terms of mass of the planet and its centripetalacceleration:Fnet = mpacWhere Fis the gravitational force, m, is the mass of the planet and a, is the centripetalacceleration of the planet.

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3. The Moon has a period of 27.3 days and a mean distance of 3.90×105 km from the center of Earth. a. Use Kepler's laws to find the period of a satellite in orbit 6.70×10³ km from the center of Earth. b. How far above Earth's surface is this satellite?

3. The Moon has a period of 27.3 days and a mean distance of 3.90×105 km from the center of Earth. a. Use Kepler's laws to find the period of a satellite in orbit 6.70×10³ km from the center of Earth. b. How far above Earth's surface is this satellite?

College Physics

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ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Question

3. The Moon has a period of 27.3 days and a mean distance of 3.90×105 km from
the center of Earth.
a. Use Kepler's laws to find the period of a satellite in orbit 6.70x103 km
from the center of Earth.
b. How far above Earth's surface is this satellite?

Law of Harmonies:
The periods of the planets are proportional to the 3/2 powers of
the major axis lengths of their orbits.
To explain further, the law states that the square of the time it takes for the planet
to make one revolution around the sun (known as the orbital period) is
proportional to the cube of the semimajor axis of the planet's elliptical orbit.
where,
T= orbital periods of a planet (planet 1 or 2)
a = length of the semimajor axis
Universal Gravitation and Kepler's Third Law
We can associate Newton's law of universal gravitation to Kepler's law of harmonies
since we can apply these equations to the motion of planets about the Sun.
Suppose a planet is orbiting the Sun. Using Newton's second law of motion, we have:
Fnet = ma
We can rewrite the equation in terms of mass of the planet and its centripetal
acceleration:
Fnet = mpac
Where Fis the gravitational force, m, is the mass of the planet and a, is the centripetal
acceleration of the planet.

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