/*! 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} Problem 15 If two satellites of different m... [FREE SOLUTION] | 91Ó°ÊÓ

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If two satellites of different masses are revolving in the same orbit, they have the same (1) angular momentum (2) energy (3) time period (4) speed

Short Answer

Expert verified
Satellites in the same orbit have the same time period and speed.

Step by step solution

01

Understanding the Problem

We are given two satellites with different masses revolving in the same orbit, and we need to determine which of the given quantities — angular momentum, energy, time period, or speed — remain the same for both satellites.
02

Identification of Relevant Concepts

For satellites revolving in a stable orbit, according to Kepler's laws of planetary motion, all objects in the same orbit will have the same orbital speed and period, regardless of their masses.
03

Applying Kepler's Third Law

Kepler's Third Law states that the time period of orbit (T) depends only on the semi-major axis of the orbit and the standard gravitational parameter of the central body, not on the mass of the revolving body. Thus, all satellites in the same orbit have the same time period.
04

Analyzing Orbital Speed

For satellites in circular orbit, the orbital speed (v) is given by the formula \[v = \sqrt{\frac{GM}{r}}\]where G is the gravitational constant, M is the mass of the central body, and r is the radius of the orbit. Hence, any satellite in the same orbit will have the same speed, regardless of its mass.

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Key Concepts

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

Orbital Speed
When it comes to the orbital dynamics of satellites, orbital speed is a crucial factor. It determines how fast a satellite travels as it circles its central body, like Earth. The formula for calculating this speed in a circular orbit is
  • \[v = \sqrt{\frac{GM}{r}}\]
Here,
  • \(G\) is the universal gravitational constant, which is approximately \(6.674 \times 10^{-11} \text{m}^3\text{kg}^{-1}\text{s}^{-2}\).
  • \(M\) is the mass of the central body, such as a planet.
  • \(r\) is the radius or the distance from the center of the central body to the satellite.
This formula shows that the mass of the satellite does not affect its speed. The speed is influenced by the distance from the center of the central body and the mass of that body. This means that any two satellites within the same orbit, regardless of their masses, will have the same orbital speed.
Time Period of Orbit
The period of orbit refers to the time a satellite takes to complete one full revolution around its primary body. According to Kepler's Third Law, this period does not depend on the satellite's mass. Instead, it is determined by
  • the semi-major axis (the average distance between the primary body and the satellite).
  • the gravitational parameter, which is the product of the gravitational constant and the central body's mass.
This leads us to the formula
  • \[T = 2\pi \sqrt{\frac{a^3}{GM}}\]
Where
  • \(T\) is the time period.
  • \(a\) represents the semi-major axis.
  • \(G\) is again the gravitational constant.
  • \(M\) is the mass of the central body.
Thus, for satellites in a stable orbit with the same semi-major axis, they all share the same time period, regardless of the satellites' own masses.
Gravitational Parameter
The gravitational parameter is a critical concept in understanding celestial mechanics. It is found by multiplying the gravitational constant \(G\) with the mass of the central body \(M\). Often represented as \(\mu\), it simplifies many orbital calculations. The equation is:
  • \[\mu = GM\]
This parameter is particularly significant when discussing orbits because it incorporates both the gravitational influence of a body and the fundamental gravitational interaction constant. In calculations such as those determining the orbital speed or time period, \(\mu\) is a key factor. The value of the gravitational parameter remains constant for a given central body, like Earth or the Sun. Therefore, it helps maintain consistent calculations for any satellites orbiting that body, ensuring the precision needed in orbital dynamics. Moreover, it simplifies the equations by combining two constants into one, making it easier to compute various orbital parameters.

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Most popular questions from this chapter

A small mass \(m\) is moved slowly from the surface of the earth to a height \(h\) above the surface. The work done (by an external agent) in doing this is (1) \(m g h\), for all values of \(h\) (2) \(m g h\), for \(h \ll R\) (3) \(1 / 2 m g R\), for \(h=R\) (4) \(-1 / 2 m g R\), for \(h=R\)

The density of a newly discovered planet is twice that of the earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of the earth. If the radius of the earth is \(R\) and the radius of the planet is \(R^{\prime}\). Then wat is the value of \(R / R^{\prime}\) ?

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