A 3000-kg satellite orbits the Earth in a circular orbit 11797 km above the Earth'ssurface (Earth radius = 6380 km, Earth Mass = 5.97x1024 kg).What is the gravitational force (in newtons, N) between the satellite and the Earth?Hint: The radius of the Earth + the height of the orbit = the center-to-center distance neededfor the equation. You also need the universal gravitational constant (G), which is not 9.81m/s2. Be careful.Gmim2F

The gravitational force between two masses m1, and m2 seperated by a distance r is given…

Answered: A 3000-kg satellite orbits the Earth in…

Glencoe Physics: Principles and Problems, Student Edition

1st Edition

ISBN:9780078807213

Author:Paul W. Zitzewitz

Publisher:Paul W. Zitzewitz

Chapter7: Gravitation

Section: Chapter Questions

Problem 67A

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Similar questions

Calculate the effective gravitational field vector g at Earths surface at the poles and the equator. Take account of the difference in the equatorial (6378 km) and polar (6357 km) radius as well as the centrifugal force. How well does the result agree with the difference calculated with the result g = 9.780356[1 + 0.0052885 sin 2 0.0000059 sin2(2)]m/s2 where is the latitude?
On a planet whose radius is 1.2107m , the acceleration due to gravity is 18m/s2 . What is the mass of the planet?
Let gM represent the difference in the gravitational fields produced by the Moon at the points on the Earths surface nearest to and farthest from the Moon. Find the fraction gM/g, where g is the Earths gravitational field. (This difference is responsible for the occurrence of the lunar tides on the Earth.)
What is the orbital radius of an Earth satellite having a period of 1.00 h? (b) What is unreasonable about this result?
The Sun has a mass of approximately 1.99 1030 kg. a. Given that the Earth is on average about 1.50 1011 m from the Sun, what is the magnitude of the Suns gravitational field at this distance? b. Sketch the magnitude of the gravitational field due to the Sun as a function of distance from the Sun. Indicate the Earths position on your graph. Assume the radius of the Sun is 7.00 108 m and begin the graph there. c. Given that the mass of the Earth is 5.97 1024 kg, what is the magnitude of the gravitational force on the Earth due to the Sun?
What is the gravitational acceleration close to the surface of a planet with a mass of 2ME and radius of 2RE where ME, and RE are the mass and radius of Earth, respectively? Answer as a multiple of g, the magnitude of the gravitational acceleration near Earths surface. (See Section 7.5.)
Suppose the gravitational acceleration at the surface of a certain moon A of Jupiter is 2 m/s2. Moon B has twice the mass and twice the radius of moon A. What is the gravitational acceleration at its surface? Neglect the gravitational acceleration due to Jupiter, (a) 8 m/s2 (b) 4 m/s2 (c) 2 m/s2 (d) 1 m/s2 (e) 0.5 m/s2
Model the Moons orbit around the Earth as an ellipse with the Earth at one focus. The Moons farthest distance (apogee) from the center of the Earth is rA = 4.05 108 m, and its closest distance (perigee) is rP = 3.63 108 m. a. Calculate the semimajor axis of the Moons orbit. b. How far is the Earth from the center of the Moons elliptical orbit? c. Use a scale such as 1 cm 108 m to sketch the EarthMoon system at apogee and at perigee and the Moons orbit. (The semiminor axis of the Moons orbit is roughly b = 3.84 108 m.)
Check Your Understanding Earth exerts a tidal force on the Moon. Is it greater than, the same as, or less than that of the Moon on Earth? Be careful in your response, as tidal forces arise from the difference in gravitational forces between one side and the other. Look at the calculations we performed for the tidal force on Earth and consider the values that would change significantly for the Moon. The diameter of the Moon is one-fourth that of Earth. Tidal forces on the Moon are not easy to detect, since there is no liquid on the surface.
Suppose astronomers find an earthlike planet that is twice the size of Earth (that is, its radius is twice that of Earth’s). What must be the mass of this planet such that the gravitational force (Fgravity) at the surface would be identical to Earth’s?
A planet has two moons with identical mass. Moon 1 is in a circular orbit of radius r. Moon 2 is in a circular orbit of radius 2r. The magnitude of the gravitational force exerted by the planet on Moon 2 is (a) four times as large (b) twice as large (c) the same (d) half as large (e) one-fourth as large as the gravitational force exerted by the planet on Moon 1.
How much work is required to lift a 500-kg satellite to an altitude of 1 · 10° m above the surface of the earth? The gravitational force is F = GMm/r², where M is the mass of the earth, m is the mass of the satellite, and r is the distance between them. The radius of the earth is 6.4 · 10° m, its mass is 6 · 10ª kg, %3D and in these units the gravitational constant, G, is 6.67 · 10~1,

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