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Chapter 6: Problem 773

orbital velocity of earth's satellite near the surface is $7 \mathrm{kms}^{-1}$. when the radius of orbit is 4 times that of earth's radius, then orbital velocity in that orbit is $=\ldots \ldots \ldots \ldots \mathrm{kms}^{-1}$ (A) \(3.5\) (B) 17 (C) 14 (D) 35

Short Answer

Expert verified
The orbital velocity of the satellite when the orbit radius is 4 times Earth's radius is (A) \(3.5\, km/s\).

Step by step solution

01

Recall the formula for orbital velocity

The formula for orbital velocity of a satellite is given by: \[v = \sqrt{\frac{GM}{r}}\] where \(v\) is the orbital velocity, \(G\) is the gravitational constant, \(M\) is the mass of the Earth, and \(r\) is the radius of the orbit.
02

Set up the given orbital velocity

The given orbital velocity near Earth's surface (\(r = R_{earth}\)) is \(7\, km/s\). We can plug this information into the formula from Step 1: \(7 = \sqrt{\frac{GM}{R_{earth}}}\)
03

Find the orbital velocity at 4 times Earth's radius

Now, we need to find the orbital velocity when the orbit radius is 4 times Earth's radius. Let this orbital velocity be \(v'\), then, we have: \(v' = \sqrt{\frac{GM}{4 \times R_{earth}}}\)
04

Find the ratio of the two orbital velocities

To find the relationship between the two orbital velocities, we can take the ratio: \(\frac{v'}{7} = \frac{\sqrt{\frac{GM}{4 \times R_{earth}}}}{\sqrt{\frac{GM}{R_{earth}}}}\)
05

Simplify the ratio

Simplify the ratio: \(\frac{v'}{7} = \frac{\sqrt{\frac{1}{4}}}{\sqrt{1}}\)
06

Solve for the new orbital velocity

Now we can solve for the new orbital velocity, \(v'\): \(v' = 7 \times \frac{1}{2}\) Therefore, the orbital velocity of the satellite when the orbit radius is 4 times Earth's radius is: \(v' = 3.5\, km/s\) So, the correct answer is: (A) \(3.5\, km/s\)

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Most popular questions from this chapter

Two bodies of masses \(m_{1}\) and \(m_{2}\) are initially at rest at infinite distance apart. They are then allowed to move towards each other under mutual a gravitational attraction Their relative velocity of approach at separation distance \(\mathrm{r}\) between them is (A) $\left[\left\\{2 \mathrm{G}\left(\mathrm{m}_{1}-\mathrm{m}_{2}\right)\right\\} / \mathrm{r}\right]^{-1 / 2}$ (B) $\left[\left\\{2 \mathrm{G}\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right)\right\\} / \mathrm{r}\right]^{1 / 2}$ (C) $\left[\mathrm{r} /\left\\{2 \mathrm{G}\left(\mathrm{m}_{1} \mathrm{~m}_{2}\right)\right\\} / \mathrm{r}\right]^{1 / 2}$ (D) $\left[\left(2 \mathrm{Gm}_{1} \mathrm{~m}_{2}\right) / \mathrm{r}\right]^{1 / 2}$

If \(\mathrm{g}\) is the acceleration due to gravity at the earth's surface and \(\mathrm{r}\) is the radius of the earth, the escape velocity for the body to escape out of earth's gravitational field is \(\ldots \ldots \ldots\) (A) \(\mathrm{gr}\) (B) \(\sqrt{(2 \mathrm{gr})}\) (C) \(\mathrm{g} / \mathrm{r}\) (D) \(\mathrm{r} / \mathrm{g}\)

The earth revolves round the sun in one year. If distance between then becomes double the new period will be years. (A) \(0.5\) (B) \(2 \sqrt{2}\) (C) 4 (D) 8

Direction (Read the following questions and choose) (A) If both Assertion and Reason are true and the Reason is correct explanation of assertion (B) If both Assertion and Reason are true, but reason is not correct explanation of the Assertion (C) If Assertion is true, but the Reason is false (D) If Assertion is false, but the Reason is true Assertion: The value of acc. due to gravity \((\mathrm{g})\) does not depend upon mass of the body Reason: This follows from \(\mathrm{g}=\left[(\mathrm{GM}) / \mathrm{R}^{2}\right]\), where \(\mathrm{M}\) is mass of planet (earth) and \(\mathrm{R}\) is radius of planet (earth) (a) \(\mathrm{A}\) (b) \(\mathrm{B}\) (c) \(\mathrm{C}\) (d) D

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