/*! 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 37 A satellite is orbiting a planet... [FREE SOLUTION] | 91影视

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Chapter 9: Problem 37

A satellite is orbiting a planet at a constant height in a circular orbit. If the mass of the planet is reduced to half, the satellite would : (a) fall on the planet (b) go to an orbit of smaller radius (c) go to an orbit of higher radius (d) escape from the planet

Short Answer

Expert verified
The satellite would go to an orbit of higher radius (option c).

Step by step solution

01

Understanding Gravitational Force

The gravitational force between the satellite and the planet is determined by the formula \( F = \frac{G m_p m_s}{r^2} \), where \( G \) is the gravitational constant, \( m_p \) is the mass of the planet, \( m_s \) is the mass of the satellite, and \( r \) is the radius of the orbit.
02

Determine Effect on Orbital Speed

For an object in circular orbit, the gravitational force provides the necessary centripetal force, \( F = \frac{m_s v^2}{r} \). When the mass of the planet is halved, the gravitational force decreases, reducing the centripetal force required for the satellite to maintain its orbit.
03

Analyze the Effect of Halving the Planet's Mass

If the mass of the planet is reduced to half, the gravitational force becomes \( F' = \frac{G (m_p/2) m_s}{r^2} \). To maintain the same gravitational pull, the satellite needs to decrease its orbital speed.
04

Consequence on Orbital Radius

Since the satellite's speed decreases, its centripetal force is reduced, causing it to move outward to a larger orbit where its new speed can still satisfy the centripetal force requirement. As the radius increases, the gravitational force aligns with this new force requirement.
05

Concluding the Effect

As the satellite moves to a higher orbit, it aligns with the new gravitational force exerted by the reduced mass of the planet, fulfilling the balance between gravitational force and required centripetal force at a larger radius.

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Key Concepts

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

Orbital Mechanics
Orbital mechanics is the branch of physics that deals with the motion of objects in space under the influence of gravity. Satellites orbit around a planet due to the gravitational force exerted on them, which acts as the centripetal force to keep them in a stable path. When a satellite orbits a planet in a circular motion, two main forces are at play:
  • Gravitational force from the planet pulling the satellite towards it.
  • Centripetal force required to keep the satellite moving in its circular path.
Changes in these forces affect the satellite's orbit. For example, altering the planet鈥檚 mass directly impacts the gravitational force, influencing whether the satellite spirals inward or outward, or maintains its current orbit. Understanding how these principles work can help in predicting satellite behavior when conditions change.
Gravitational Force
The gravitational force is what keeps satellites in orbit around planets. It is a central force that acts between any two masses, attracting each towards the other. The strength of this force is calculated using the formula:\[ F = \frac{G m_1 m_2}{r^2} \]where:
  • \( G \) = Universal Gravitation Constant
  • \( m_1 \) and \( m_2 \) = Masses of the two interacting objects
  • \( r \) = Distance between the centers of two objects
In the context of our exercise, reducing the planet鈥檚 mass by half decreases the gravitational force by half, affecting the satellite's motion. A lower gravitational pull would not provide enough centripetal force to maintain the satellite in its original orbit, prompting a transition to a different orbital path.
Centripetal Force
Centripetal force is the necessary inward force that keeps an object moving in a circular path. For a satellite in orbit, the gravitational force from the planet acts as this centripetal force. Calculated by the formula:\[ F = \frac{m v^2}{r} \]where:
  • \( m \) = Mass of the satellite
  • \( v \) = Orbital speed of the satellite
  • \( r \) = Radius of the orbit
As the mass of the planet changes, the gravitational force (providing centripetal force) changes. If the planet's mass is halved, the gravitational pull weakens, hence the satellite鈥檚 orbital speed decreases. Without the proper centripetal force, the satellite moves to a higher orbit where its speed and the reduced gravitational force are balanced, aligning with new force requirements to sustain orbiting stability.

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Most popular questions from this chapter

Three point masses each of mass \(m\) rotate in a circle of radius \(r\) with constant angular velocity \(\omega\) due to their mutual gravitational attraction. If at any instant, the masses are on the vertex of an equilateral triangle of side \(a\), then the value of \(\omega\) is: (a) \(\sqrt{\frac{G m}{a^{3}}}\) (b) \(\sqrt{\frac{3 G m}{a^{3}}}\) (c) \(\sqrt{\frac{G m}{3 a^{3}}}\) (d) none

Two satellites \(S\) and \(S^{\prime}\) revolve around the earth at distances \(3 R\) and \(6 R\) from the centre of earth. Their periods of revolution will be in the ratio: (a) \(1: 2\) (b) \(2: 1\) (c) \(1: 2^{1.5}\) (d) \(1: 2^{0.67}\)

A point mass \(m_{0}\) is placed at distance \(R / 3\) from the centre of a spherical shell of mass \(m\) and radius \(R\). The gravitational force on the point mass \(m_{0}\) is: (a) \(\frac{4 G m m_{0}}{R^{2}}\) (b) zero (c) \(\frac{9 G m m_{0}}{R^{2}}\) (d) none of these

A straight rod of length \(L\) extends from \(x=a\) to \(x=L+a\). The gravitational force exerted by it on a point mass \(m\) at \(x=0\) if the linear density of rod is \(\mu=A+B x^{2}\), is: (a) \(\mathrm{Gm}\left[\frac{A}{a}+B L\right]\) (b) \(G m\left[A\left(\frac{1}{a}-\frac{1}{a+L}\right)+B L\right]\) (c) \(\mathrm{Gm}\left[B L+\frac{A}{a+L}\right]\) (d) \(G m\left[B L-\frac{A}{a}\right]\)

Two bodies each of mass \(1 \mathrm{~kg}\) are at a distance of \(1 \mathrm{~m}\). The escape velocity of a body of mass \(1 \mathrm{~kg}\) which is midway between them is: (a) \(8 \times 10^{-5} \mathrm{~m} / \mathrm{s}\) (b) \(2.31 \times 10^{-5} \mathrm{~m} / \mathrm{s}\) (c) \(4.2 \times 10^{-5} \mathrm{~m} / \mathrm{s}\) (d) zero
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