21st Century Astronomy
6th Edition
ISBN: 9780393428063
Author: Kay
Publisher: NORTON
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Chapter 4, Problem 35QP
To determine
The time takes for Newton’s cannonball to complete one orbit around Earth.
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If a satellite orbits the Earth at a height of 5,642.74 km above the Earth's surface what is its orbital period in unit hours? Earth's radius is 6370 km, Earth's mass is 5.98 x1024 kg, and G=6.67x 10 -11 Nm2/kg2.
A 9.0 × 10 3 kg satellite with an orbital radius of 3.20 × 10 7 m orbits the earth at an altitude of 2.56 × 10 7 m. What is the orbital period?
The moon of a certain planet is observed to have a nearly circular orbit of radius 3.89∗10^5 km, and an orbital period of 25.4 days. What is the mass of the planet?
7.23∗10^24 kg
2.32∗10^24 kg
4.89∗10^25 kg
4.65∗10^25 kg
Chapter 4 Solutions
21st Century Astronomy
Ch. 4.1 - Prob. 4.1ACYUCh. 4.1 - Prob. 4.1BCYUCh. 4.2 - Prob. 4.2CYUCh. 4.3 - Prob. 4.3CYUCh. 4.4 - Prob. 4.4CYUCh. 4 - Prob. 1QPCh. 4 - Prob. 2QPCh. 4 - Prob. 3QPCh. 4 - Prob. 4QPCh. 4 - Prob. 5QP
Ch. 4 - Prob. 6QPCh. 4 - Prob. 7QPCh. 4 - Prob. 8QPCh. 4 - Prob. 9QPCh. 4 - Prob. 10QPCh. 4 - Prob. 11QPCh. 4 - Prob. 12QPCh. 4 - Prob. 13QPCh. 4 - Prob. 14QPCh. 4 - Prob. 15QPCh. 4 - Prob. 16QPCh. 4 - Prob. 17QPCh. 4 - Prob. 18QPCh. 4 - Prob. 19QPCh. 4 - Prob. 20QPCh. 4 - Prob. 21QPCh. 4 - Prob. 22QPCh. 4 - Prob. 23QPCh. 4 - Prob. 24QPCh. 4 - Prob. 25QPCh. 4 - Prob. 26QPCh. 4 - Prob. 27QPCh. 4 - Prob. 28QPCh. 4 - Prob. 29QPCh. 4 - Prob. 30QPCh. 4 - Prob. 31QPCh. 4 - Prob. 32QPCh. 4 - Prob. 33QPCh. 4 - Prob. 34QPCh. 4 - Prob. 35QPCh. 4 - Prob. 36QPCh. 4 - Prob. 37QPCh. 4 - Prob. 38QPCh. 4 - Prob. 39QPCh. 4 - Prob. 40QPCh. 4 - Prob. 41QPCh. 4 - Prob. 42QPCh. 4 - Prob. 43QPCh. 4 - Prob. 44QPCh. 4 - Prob. 45QP
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- Eros has an elliptical orbit about the Sun, with a perihelion distance of 1.13 AU and aphelion distance of 1.78 AU. What is the period of its orbit?arrow_forwardIt was stated that a satellite with negative total energy is in a bound orbit, whereas one with zero or positive total energy is in an unbounded orbit. Why zero or positive total energy is in an unbounded orbit. Why is this true? What choice for gravitational potential energy was made such that this is true?arrow_forwardUse the following information to help determine the orbital velocity and orbital period of a satellite at a location of 15000 miles above the surface of the earth. Mass of Earth = 5.98 x 1024 kgRadius of Earth = 6.37 x 106 m1609 m = 1.00 miarrow_forward
- The gravitational force between a planet and its moon is 6*10^16 N. For another moon orbiting the same planet with twice the orbital radius and 3 times the mass, what is the gravitational force?arrow_forwardA planet of mass 6 x 1024 kg is in orbit around a star of mass 2 x 1030 kg. If the distance between the planet and the star is 5 x 1011 m, what is the force exerted on this planet by the star?arrow_forwardif the average speed of a planet in its orbit is 35 km/hr what would be the speed of the planet when it is closest to the sun? A.35 km/hr B.25 km/hr C.20 km/hr 45. 45 km/hrarrow_forward
- If a satellite circulate around the earth at a height of 2,317.84 km above the earth's surface, given the earth radius is 3958.8 miles and mass is 5.98 x1024 kg, use G=6.67x 10 -11 Nm2/kg2, find the period of this satellite in unit hours?arrow_forwardIf a satellite with a mass of 9 x 10^2 kg is placed in an orbit around a planet with a mass of 8 x 10^20 kg, at a speed of 19000 m/s, find the distance from the center of the planet to the satellite.arrow_forwardWhich scientist is credited with the collection of data necessary to support the planet's elliptical motion?arrow_forward
- Earth’s mass is 5.98 x 1024 kg and it is located 150 million kilometers from the Sun. Calculate the mass of the Sun in [x1030 kg]. Find the speed, in [m/s], of a satellite orbiting the Earth 1000 km above its surface. (RE = 6.37 x 103 km & ME = 5.98 x 1024 kg) Find the period, in [s], of a satellite orbiting the Earth 1000 km above its surface. (RE = 6.37 x 103 km & ME = 5.98 x 1024 kg)arrow_forwardThe orbital speeds of the planets Mercury and Mars are v(Mercury) = 50 km / s and V(Mars) = 24.2 km / s. Assume that the orbits of these planets Mercury and Mars around the Sun are circular, find the ratio of the radii of their R (Mercury) / R (Mars) orbits.arrow_forwardWhy does a satellite need no fuel to move round a planet in a fixed orbit?arrow_forward
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