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Chapter 6: Problem 783

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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Most popular questions from this chapter

A satellite of mass \(\mathrm{m}\) is circulating around the earth with constant angular velocity. If radius of the orbit is \(\mathrm{R}_{0}\) and mass of earth \(\mathrm{M}\), the angular momentum about the center of earth is (A) \(m \sqrt{\left(G M R_{0}\right)}\) (B) \(\mathrm{M} \sqrt{\left(\mathrm{GMR}_{\mathrm{o}}\right)}\) (C) \(\left.m \sqrt{(G M} / R_{0}\right)\) (D) \(\mathrm{M} \sqrt{\left(\mathrm{GM} / \mathrm{R}_{\mathrm{o}}\right)}\)

A particle of mass \(\mathrm{M}\) is situated at the center of a spherical shell of same mass and radius a the magnitude of gravitational potential at a point situated at (a/2) distance from the center will be (A) \([(4 \mathrm{GM}) / \mathrm{a}]\) (B) \((\mathrm{GM} / \mathrm{a})\) (C) \([(2 \mathrm{GM}) / \mathrm{a}]\) (D) \([(3 \mathrm{GM}) / \mathrm{a}]\)

Two small and heavy sphere, each of mass \(\mathrm{M}\), are placed distance r apart on a horizontal surface the gravitational potential at a mid point on the line joining the center of spheres is (A) zero (B) \(-(\mathrm{GM} / \mathrm{r})\) (C) \(-[(2 \mathrm{GM}) / \mathrm{r}]\) (D) \(-[(4 \mathrm{GM}) / \mathrm{r}]\)

The moon's radius is \(1 / 4\) that of earth and its mass is \(1 / 80\) times that of the earth. If g represents the acceleration due to gravity on the surface of earth, that on the surface of the moon is (A) \(g / 4\) (B) \(\mathrm{g} / 5\) (c) \(\mathrm{g} / 6\) (D) \(\mathrm{g} / 8\)

The escape velocity of an object from the earth depends upon the mass of earth (M), its mean density ( \(p\) ), its radius (R) and gravitational constant (G), thus the formula for escape velocity is (A) \(U=\mathrm{R} \sqrt{[}(8 \pi / 3) \mathrm{Gp}]\) (C) \(\mathrm{U}=\sqrt{(2 \mathrm{GMR})}\) (D) \(U=\sqrt{\left[(2 \mathrm{GMR}) / \mathrm{R}^{2}\right]}\)
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