Chapter 7: Problem 6
The period of a satellite in a circular orbit around a planet is independent of, (A) the mass of the planet. (B) the radius of the planet. (C) the mass of the satellite. (D) all of three parameters \(a, b\) and \(c\).
/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none}
Learning Materials
Features
Discover
Chapter 7: Problem 6
The period of a satellite in a circular orbit around a planet is independent of, (A) the mass of the planet. (B) the radius of the planet. (C) the mass of the satellite. (D) all of three parameters \(a, b\) and \(c\).
All the tools & learning materials you need for study success - in one app.
Get started for free
The mass of a spaceship is \(1000 \mathrm{~kg}\). It is to be launched from the earth's surface out into free space. The value of \(g\) and \(r\) (radius of earth) are \(10 \mathrm{~m} / \mathrm{s}^{2}\) and \(6400 \mathrm{~km}\) respectively. The required energy for this work will be (A) \(6.4 \times 10^{11} \mathrm{~J}\) (B) \(6.4 \times 10^{8} \mathrm{~J}\) (C) \(6.4 \times 10^{9} \mathrm{~J}\) (D) \(6.4 \times 10^{10} \mathrm{~J}\)
Four particles, each of mass \(M\) and equidistant from each other, move along a circle of radius \(R\) under the action of their mutual gravitational attraction. The speed of each particle is (A) \(\sqrt{\frac{G M}{R}}\) (B) \(\sqrt{2 \sqrt{2} \frac{G M}{R}}\) (C) \(\sqrt{\frac{G M}{R}(1+2 \sqrt{2})}\) (D) \(\frac{1}{2} \sqrt{\frac{G M}{R}(1+2 \sqrt{2})}\)
A planet in a distance solar system is 10 times more massive than the earth and its radius is 10 times smaller. Given that the escape velocity from the earth is \(11 \mathrm{kms}^{-1}\), the escape velocity from the surface of the planet would be. (A) \(1.1 \mathrm{kms}^{-1}\) (B) \(11 \mathrm{kms}^{-1}\) (C) \(110 \mathrm{kms}^{-1}\) (D) \(0.11 \mathrm{kms}^{-1}\)
The height at which the acceleration due to gravity becomes \(\frac{g}{9}\) (where \(g\) = the acceleration due to gravity on the surface of the earth) in terms of \(R\), the radius of the earth is (A) \(\frac{R}{\sqrt{2}}\) (B) \(R / 2\) (C) \(\sqrt{2} R\) (D) \(2 R\)
A planet is revolving around the sun in elliptical orbit. Its closest distance from the sun is \(r\) and the farthest distance is \(R\). If the orbital velocity of the planet closest to the sun be \(v\), then what is the velocity at the farthest point? (A) \(\overline{v r} / R\) (B) \(v R / r\) (C) \(v\left(\frac{r}{R}\right)^{1 / 2}\) (D) \(v\left(\frac{R}{r}\right)^{1 / 2}\)
What do you think about this solution?
We value your feedback to improve our textbook solutions.