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Chapter 10: Problem 24

If satellite is revolving around a planet of mass \(M\) in an elliptical orbit of semi-major axis \(a\), find the orbital speed of the satellite when it is at a distance \(r\) from the focus. (a) \(v^{2}=G M\left[\frac{2}{r}-\frac{1}{a}\right]\) (b) \(v^{2}=G M\left[\frac{2}{r^{2}}-\frac{1}{a}\right]\) (c) \(v^{2}=G M\left[\frac{2}{r^{2}}-\frac{1}{a^{2}}\right]\) (d) \(v^{2}=G\left[\frac{2}{r}-\frac{1}{a}\right]\)

Short Answer

Expert verified
The correct answer is (a) \( v^{2}=G M\left[\frac{2}{r}-\frac{1}{a}\right] \).

Step by step solution

01

Understanding the Problem

We need to determine the velocity of a satellite moving in an elliptical orbit at a distance r from the focus of the planet's mass. The formula for the speed of a satellite in an elliptical orbit can be derived from the vis-viva equation: \[ v^2 = GM \left( \frac{2}{r} - \frac{1}{a} \right) \] where G is the gravitational constant, M is the mass of the planet, r is the current distance of the satellite from the focus, and a is the semi-major axis of the orbit.
02

Analyzing Given Options

We will compare each given answer to the vis-viva equation \[ v^2 = GM \left( \frac{2}{r} - \frac{1}{a} \right) \]. Let's examine each option: - (a) matches the derived vis-viva equation perfectly.- (b) has an incorrect term \( \frac{2}{r^2} \) instead of \( \frac{2}{r} \).- (c) also incorrectly uses \( \frac{1}{a^2} \) instead of \( \frac{1}{a} \).- (d) lacks the mass term M in the expression, making it incorrect.
03

Verifying the Correct Formula

The vis-viva equation calculated in Step 1 is \( v^2 = GM \left( \frac{2}{r} - \frac{1}{a} \right) \). This matches option (a) exactly, which confirms that (a) is the correct answer according to both theoretical knowledge and the given problem.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Elliptical orbits
Elliptical orbits are the paths followed by celestial bodies, such as planets, satellites, or comets, under the influence of gravity.
These paths are not perfect circles, but elongated ellipses, much like an oval shape. The key characteristics of an ellipse include:
  • Semi-major axis: the longest radius of the ellipse, extending from its center to the edge.
  • Foci: two fixed points inside the ellipse; in an orbital context, one focus is often occupied by the central massive body, like the Sun or a planet.
These features help scientists and engineers calculate the path and motion of satellites and other orbiting bodies accurately.
A satellite's speed varies along its elliptical orbit, being greatest when near the closest point to the massive body (perigee) and lowest when at the farthest point (apogee).
Orbital speed
Orbital speed is the velocity at which a satellite or any celestial body travels along its orbit.
In an elliptical orbit, this speed is not constant but varies depending on the satellite's position.The vis-viva equation is particularly useful in determining the orbital speed at any given point in an elliptical orbit:
\[ v^2 = GM \left( \frac{2}{r} - \frac{1}{a} \right) \]Here's what the terms mean:
  • \(v\): orbital speed of the satellite.
  • \(G\): gravitational constant, a universal value that affects all matter.
  • \(M\): mass of the central body being orbited.
  • \(r\): current distance of the satellite from the focal point, where the planet's mass is concentrated.
  • \(a\): semi-major axis of the orbit.
The above equation shows that orbital speed is greater when the satellite is closer to the planet due to gravity's pull and decreases as the distance increases.
Gravitational constant
The gravitational constant, often denoted by \(G\), is a fundamental constant in physics that quantifies the strength of gravitational force between two masses.
Its value is approximately \(6.674 \times 10^{-11} \text{ m}^3 \text{ kg}^{-1} \text{ s}^{-2}\).This constant is a critical component of Newton's law of gravitation, which describes the attraction force between any two bodies in the universe:
\[ F = \frac{GMm}{r^2} \]where:
  • \(F\): gravitational force between the two masses.
  • \(M\) and \(m\): masses of the two objects.
  • \(r\): distance between the centers of the two masses.
By incorporating \(G\), we can predict and calculate gravitational effects on objects and understand phenomena like orbits and tides.
It helps establish the fundamental linkage between mass, distance, and gravitational attraction.

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Most popular questions from this chapter

The gravitational attraction between the two bodies increases when their masses are (a) reduced and distance is reduced (b) increased and distance is reduced (c) reduced and distance is increased (d) increased and distance is increased

The ratio of acceleration due to gravity at a height \(h\) above the surface of the earth and at a depth \(h\) below the surface of the earth for \(h<\) radius of earth (a) is constant (b) increases linearly with \(h\) (c) decreases linearly with \(h\) (d) decreases parabolically with \(h\)

Two equal mases \(m\) and \(m\) are hung from balance whose scale pans differ in vertical height by \(h\). Calculate the error in weighing, if any, in terms of density of earth \(\rho\). (a) \(\frac{2}{3} \pi \rho R^{3} \mathrm{Gm}\) (b) \(\frac{8}{3} \pi \rho G m h\) (c) \(\frac{8}{3} \pi \rho R^{3} G m\) (d) \(\frac{4}{3} \pi \rho G m^{2} h\)

A satellite orbits the earth at a height of \(400 \mathrm{~km}\) above the surface. How much energy must be expended to rocket the satellite out of the earth's gravitational influence? Mass of the satellite \(=200 \mathrm{~kg}\), mass of the earth \(=6.0 \times 10^{24} \mathrm{~kg}\), radius of the earth \(=6.4 \times 10^{6} \mathrm{~m}, G=6.67 \times 10^{-11} \mathrm{~N}-\mathrm{m}^{2} / \mathrm{kg}^{2}\). (a) \(5.2 \times 10^{10} \mathrm{~J}\) (b) \(3 \times 10^{6} \mathrm{~J}\) (c) \(4 \times 10^{6} \mathrm{~J}\) (d) \(6 \times 10^{9} \mathrm{~J}\)

Linear acceleration for rolling, \(a=\frac{g \sin \theta}{\sqrt{1+\frac{K^{2}}{R^{2}}}}\) For cylinder, \(\frac{K^{2}}{R^{2}}=\frac{1}{2}\) \(\therefore \quad a_{\text {cylinder }}=\frac{2}{3} g \sin \theta\) For rotation, the torque $$ f R=l \alpha \cdot\left(M R^{2} \alpha\right) / 2 $$ (where, \(f=\) force of friction) But \(R \alpha=a \quad \therefore f=\frac{M}{2} a\) \(\therefore \quad f=\frac{M}{2} \cdot \frac{2}{3} g \sin \theta=\frac{M}{3} g \sin \theta\) \(\mu_{s}=f / N\), where \(N\) is normal reaction, \(\therefore\) $$ \mu_{s}=\frac{\frac{M}{3} g \sin \theta}{M g \cos \theta}=\frac{\tan \theta}{3} $$ \(\therefore\) For rolling without slipping of a roller down the inclined plane, \(\tan \theta \leq 3 \mu_{s}\).
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