(a)
ToCalculate: The ratio of spin angular momenta of Mars and Earth.
(a)
Answer to Problem 21P
Explanation of Solution
Given information :
Mars and Earth have nearly identical lengths of days.
Earth’s mass is
Radius is
Mars’ orbital radius is on an average
The Martian year is
Formula used :
The
Where, I is the moment of inertia and
Moment of inertia of sphere is
Where, M is the mass and R is the radius of the sphere.
Calculation:
As Mars and Earth have nearly identical lengths of days.
The ratio of spin angular momenta of Mars and Earth is:
Conclusion:
The ratio of spin angular momenta of Mars and Earth is 33:1.
(b)
ToCalculate: The ratio of spin kinetic energies of Mars and Earth.
(b)
Answer to Problem 21P
Explanation of Solution
Given information :
Mars and Earth have nearly identical lengths of days.
Earth’s mass is
Radius is
Mars’ orbital radius ison an average
The Martian year is
Formula used :
Rotational kinetic energy is:
Where, I is the moment of inertia and
Moment of inertia of sphere is
Where, M is the mass and R is the radius of the sphere.
Calculation:
As Mars and Earth have nearly identical lengths of days.
Conclusion:
The ratio of spin kinetic energies of Mars and Earth is 33:1.
(c)
ToCalculate: The ratioorbital angular momenta of Mars and Earth.
(c)
Answer to Problem 21P
Explanation of Solution
Given information :
Mars and Earth have nearly identical lengths of days.
Earth’s mass is
Radius is
Mars’ orbital radius ison an average
The Martian year is
Formula used :
The angular momentum is given by:
Where, I is the moment of inertia and
Moment of inertia of sphere is
Where, M is the mass and R is the radius of the sphere.
Calculation:
Treating Earth and Mars as point objects, the ratio of their orbital angular momenta is
Substituting for the moments of inertia and angular speeds yields
Where
Substitute numerical values for the three ratios and evaluate
Conclusion:
The ratioorbital angular momenta of Mars and Earth is,
(d)
ToCalculate: The ratio of orbital kinetic energies of Mars and Earth.
(d)
Answer to Problem 21P
Explanation of Solution
Given information :
Mars and Earth have nearly identical lengths of days.
Earth’s mass is
Radius is
Mars’ orbital radius is, on average
The Martian year is
Formula used :
Rotational kinetic energy is:
Where, I is the moment of inertia and
Moment of inertia of sphere is
Where, M is the mass and R is the radius of the sphere.
Calculation:
Conclusion:
The ration of orbital kinetic energies of Mars and Earth is
Want to see more full solutions like this?
Chapter 10 Solutions
Physics For Scientists And Engineers Student Solutions Manual, Vol. 1
- On a planet whose radius is 1.2107m , the acceleration due to gravity is 18m/s2 . What is the mass of the planet?arrow_forwardSuppose 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/s2arrow_forwardThe 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?arrow_forward
- For many years, astronomer Percival Lowell searched for a Planet X that might explain some of the perturbations observed in the orbit of Uranus. These perturbations were later explained when the masses of the outer planets and planetoids, particularly Neptune, became better measured (Voyager 2). At the time, however, Lowell had proposed the existence of a Planet X that orbited the Sun with a mean distance of 43 AU. With what period would this Planet X orbit the Sun?arrow_forwardIn Example 2.6, we considered a simple model for a rocket launched from the surface of the Earth. A better expression for the rockets position measured from the center of the Earth is given by y(t)=(R3/2+3g2Rt)2/3j where R is the radius of the Earth (6.38 106 m) and g is the constant acceleration of an object in free fall near the Earths surface (9.81 m/s2). a. Derive expressions for vy(t) and ay(t). b. Plot y(t), vy(t), and ay(t). (A spreadsheet program would be helpful.) c. When will the rocket be at y=4R? d. What are vy and ay when y=4R?arrow_forwardLet 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.)arrow_forward
- The “mean” orbital radius listed for astronomical objects orbiting the Sun is typically not an integrated average but is calculated such that it gives the correct period when applied to the equation for circular orbits. Given that, what is the mean orbital radius in terms of aphelion and perihelion?arrow_forwardCalculate the mass of the Sun based on data for average Earth’s orbit and compare the value obtained with the Sun’s commonly listed value of 1.9891030kg .arrow_forwardWhat is the orbital radius of an Earth satellite having a period of 1.00 h? (b) What is unreasonable about this result?arrow_forward
- A geosynchronous Earth satellite is one that has an orbital period of precisely 1 day. Such orbits are sueful for communication and weather observation because the satellite remains above the same point on Earth (provided it orbits in the equatorial plane in the same direction as Earth’s rotation). Calculate the radius of such an orbit based on the data for Earth in Appendis D.arrow_forwardTwo planets in circular orbits around a star have speed of v and 2v . (a) What is the ratio of the orbital radii of the planets? (b) What is the ratio of their periods?arrow_forwardThe astronaut orbiting the Earth in Figure P3.27 is preparing to dock with a Westar VI satellite. The satellite is in a circular orbit 600 km above the Earth’s surface, where the free-fall acceleration is 8.21 m/s2. Take the radius of the Earth as 6 400 km. Determine the speed of the satellite and the time interval required to complete one orbit around the Earth, which is the period of the satellite. Figure P3.27arrow_forward
- Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningPhysics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage LearningUniversity Physics Volume 1PhysicsISBN:9781938168277Author:William Moebs, Samuel J. Ling, Jeff SannyPublisher:OpenStax - Rice University
- College PhysicsPhysicsISBN:9781285737027Author:Raymond A. Serway, Chris VuillePublisher:Cengage LearningClassical Dynamics of Particles and SystemsPhysicsISBN:9780534408961Author:Stephen T. Thornton, Jerry B. MarionPublisher:Cengage LearningPhysics for Scientists and Engineers, Technology ...PhysicsISBN:9781305116399Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning