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When a satellite going round the earth in a circular orbit of radius \(\mathrm{r}\) and speed \(\mathrm{v}\) loses some of its energy, then \(\mathrm{r}\) and \(\mathrm{v}\) changes as (A) \(r\) and \(v\) both will increase (B) \(\mathrm{r}\) and \(\mathrm{v}\) both will decease (C) \(r\) will decrease and \(\mathrm{v}\) will increase (D) \(\mathrm{r}\) will increase and \(\mathrm{v}\) will decrease

Short Answer

Expert verified
When a satellite loses energy in a circular orbit, the radius (r) will decrease, and the speed (v) will increase. This is demonstrated by the conservation of angular momentum and gravitational potential energy formulas, leading to the conclusion that the correct answer is option (C).

Step by step solution

01

Write down the conservation of angular momentum formula

The conservation of angular momentum formula for a satellite orbiting in a circle can be expressed as follows: \[L = mrv\] where L is the angular momentum, m is the mass of the satellite, r is the radius of the orbit, and v is the speed of the satellite.
02

Calculate the gravitational potential energy

We can compute the gravitational potential energy of the satellite using this formula: \[U = -\frac{GMm}{r}\] where U is the gravitational potential energy, G is the gravitational constant, M is the mass of Earth, and r is the orbit radius.
03

Express v in terms of the gravitational potential energy

To find the relationship between r and v when energy is lost, we'll first express v in terms of U. Using the conservation of angular momentum formula, we have: \[v = \frac{L}{mr}\]
04

Substitute the expression for v into the gravitational potential energy equation

Now, substitute the expression for v from Step 3 into the gravitational potential energy equation: \[U = -\frac{GMm}{r} = -\frac{GML^2}{(mr)^3}\]
05

Derive the relationship between radius and speed

As the satellite loses energy (a decrease in U), the potential energy becomes less negative. Re-write the expression for U and analyze the behavior of r and v: \[\frac{GMm}{r} = \frac{GML^2}{(mr)^3}\] \[\frac{r^2}{L^2} = \frac{GM}{l}\] Since the angular momentum (L) is conserved and does not change, notice that if r decreases, the value of v must increase to maintain the balance in the equation. Therefore, we arrive at the following conclusion: (C) r will decrease and v will increase Hence, the correct answer is option (C).

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Key Concepts

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

Angular Momentum Conservation
Angular momentum conservation is a critical principle in physics, especially in the study of orbital dynamics. When dealing with satellites in orbit, this principle states that the angular momentum (L) of a satellite is conserved as long as no external torque acts on the system. For a satellite in a circular orbit, this can be understood via the formula: \[ L = mrv \]
  • Here, \(m\) is the mass of the satellite, \(r\) stands for the radius of the orbit, and \(v\) represents the orbital speed.
Since the product \(mrv\) must remain constant if the satellite's mass does not change, any change in the orbital radius \(r\) necessitates a compensatory change in speed \(v\) to maintain \(L\) constant. This inherent balance plays a critical role in understanding how satellites adjust their paths and speeds when slight changes in energy occur.
Gravitational Potential Energy
Gravitational Potential Energy (U) is another pivotal concept when discussing the motion of satellites. This energy is related to an object's position within a gravitational field and is given by the formula:\[ U = -\frac{GMm}{r} \]
  • \(G\) is the gravitational constant; \(M\) represents the Earth's mass; \(m\) stands for the satellite's mass; and \(r\) denotes the radius of the satellite's orbit.
The negative sign indicates that gravitational forces are attractive. As the satellite loses energy, variation in \(U\) leads to changes in its orbit's dimensions. The gravitational potential energy decreases in magnitude (becomes less positive) as the satellite gets closer to Earth. This seemingly simple idea forms the basis for understanding why changes in energy affect both the satellite's speed and the radius of its orbit.
Satellite Motion
Understanding satellite motion involves recognizing the intricate balance between gravitational pull and inertia. Satellites stay in orbit because of this balance. The gravitational force between the satellite and the Earth ensures that the satellite follows a curved path rather than moving off in a straight line. Concurrently, the satellite's inertia wants to keep it moving along its current path.When a satellite loses some energy, its gravitational potential energy decreases, and the satellite is pulled closer towards the Earth, decreasing the orbital radius \(r\). This change affects speed because the satellite needs to move faster to stay in orbit due to conservation laws. Essentially, the interplay of forces and principles keeps satellites stably circling, vital for everything from telecommunications to weather forecasting.
Circular Orbits
Circular orbits are a type of satellite motion where the path traced by the satellite forms a perfect circle. In such orbits, the speed \(v\) and the radius \(r\) of the path remain constant under ideal conditions. However, when external factors disturb this balance, such as a loss in energy, these values can change. The increase in speed as the orbit decreases is a direct outcome of conserving angular momentum.
  • Here, a balance must be achieved between the gravitational pull towards Earth and the satellite's momentum, pushing it tangentially away.
  • Circular orbits are well-suited for many tasks, including geostationary satellites, which need a constant position relative to Earth's surface.
A fundamental understanding of these dynamics helps in planning satellite paths, ensuring they function optimally and stay on their intended trajectory, reflecting real-world applications of these physics concepts.

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Most popular questions from this chapter

He period of revolution of planet \(\mathrm{A}\) around the sun is 8 times that of \(\mathrm{B}\). The distance of A from the sun is how many times greater than that of \(\mathrm{B}\) from the sun. (A) 2 (B) 3 (C) 4 (D) 5

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