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Chapter 6: Problem 771

A satellite is moving around the earth with speed \(\mathrm{v}\) in a circular orbit of radius \(\mathrm{r}\). If the orbit radius is decreased by \(1 \%\) its speed will (A) increase by \(1 \%\) (B) increase by \(0.5 \%\) (C) decreased by \(1 \%\) (C) Decreased by \(0.5 \%\)

Short Answer

Expert verified
The speed of the satellite will increase by approximately \(0.5\%\) when the orbit radius is decreased by \(1\%\).

Step by step solution

01

Write down the formula for gravitational force and centripetal force

The formula for gravitational force acting on an object is given by \(F_g = G \frac{Mm}{r^2}\), where G is the universal gravitational constant, M is the mass of the Earth, m is the mass of the satellite, and r is the distance between their centers. The formula for centripetal force acting on an object moving in a circle of radius r and speed v is given by \(F_c = mv^2/r\), where m is the mass of the satellite and v is the speed of the satellite.
02

Equate gravitational force and centripetal force

Since the satellite is moving in a circular orbit, the gravitational force acting on it will be equal to the centripetal force. Therefore, we can equate the two formulas: \(G \frac{Mm}{r^2} = m \frac{v^2}{r}\)
03

Solve for v in terms of r

We can cancel \(\frac{1}{r}\) on both sides and divide by m to find an expression for v^2: \(\frac{GM}{r} = v^2\) Take square root on both sides: \(v = \sqrt{\frac{GM}{r}}\)
04

Calculate the new values for r and v

When the radius of the orbit is decreased by 1%, the new radius will be 0.99r. We can use the derived expression for v to find the new speed of the satellite: \(v' = \sqrt{\frac{GM}{0.99r}}\)
05

Calculate the percentage change in speed

We can now find the percentage change in speed as follows: \(\% \Delta v = \frac{v' - v}{v} \times 100\) Substitute the expressions for v and v': \(\% \Delta v = \frac{\sqrt{\frac{GM}{0.99r}} - \sqrt{\frac{GM}{r}}}{\sqrt{\frac{GM}{r}}} \times 100\) Simplify the expression: \(\% \Delta v = \left(\frac{\sqrt{\frac{1}{0.99}} - 1}{1}\right) \times 100\)
06

Calculate the numerical value

Using a calculator, we find the percentage change in speed to be approximately: \(\% \Delta v \approx 0.503 \%\) So, the answer is (B) increase by 0.5%.

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Most popular questions from this chapter

The acceleration due to gravity near the surface of a planet of radius \(\mathrm{R}\) and density \(\mathrm{d}\) is proportional to (A) \(\mathrm{d} / \mathrm{R}^{2}\) (B) \(\mathrm{d} \mathrm{R}^{2}\) (C) \(\mathrm{dR}\) (D) \(\mathrm{d} / \mathrm{R}\)

The largest and shortest distance of the earth from the sun are \(\mathrm{r}_{1}\) and \(\mathrm{r}_{2}\) its distance from the sun when it is at the perpendicular to the major axis of the orbit drawn from the sun (A) \(\left[\left(\mathrm{r}_{1}+\mathrm{r}_{2}\right) / 4\right]\) (B) $\left[\left(\mathrm{r}_{1} \mathrm{r}_{2}\right) /\left(\mathrm{r}_{1}+\mathrm{r}_{2}\right)\right]$ (C) \(\left[\left(2 r_{1} r_{2}\right) /\left(r_{1}+r_{2}\right)\right]\) (D) \(\left[\left(\mathrm{r}_{1}+\mathrm{r}_{2}\right) / 3\right]\)

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) \(U=\sqrt{\left[(2 \mathrm{GMR}) / \mathrm{R}^{2}\right]}\)

If the value of ' \(\mathrm{g}\) ' acceleration due to gravity, at earth surface is \(10 \mathrm{~ms}^{-2}\). its value in \(\mathrm{ms}^{-2}\) at the center of earth, which is assumed to be a sphere of Radius ' \(\mathrm{R}\) 'meter and uniform density is (A) 5 (B) \(10 / \mathrm{R}\) (C) \(10 / 2 \mathrm{R}\) (D) zero

What does not change in the field of central force (A) potential energy (B) Kinetic energy (C) linear momentum (D) Angular momentum
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