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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

If the density of small planet is that of the same as that of the earth while the radius of the planet is \(0.2\) times that of the earth, the gravitational acceleration on the surface of the planet is (A) \(0.2 \mathrm{~g}\) (B) \(0.4 \mathrm{~g}\) (C) \(2 \mathrm{~g}\) (D) \(4 \mathrm{~g}\)

If \(\mathrm{V}_{\mathrm{e}}\) and \(\mathrm{V}_{\mathrm{o}}\) are represent the escape velocity and orbital velocity of satellite corresponding to a circular orbit of -adius \(\mathrm{r}\), then A) \(\mathrm{V}_{\mathrm{e}}=\mathrm{V}_{\mathrm{o}}\) (B) \(\sqrt{2} \mathrm{~V}_{\mathrm{o}}=\mathrm{V}_{\mathrm{e}}\) C) \(\mathrm{V}_{\mathrm{e}}=\left(\mathrm{V}_{\mathrm{O}} / \sqrt{2}\right)\) (D) \(\mathrm{V}_{\mathrm{e}}\) and \(\mathrm{V}_{\mathrm{o}}\) are not related

If mass of a body is \(\mathrm{M}\) on the earth surface, than the mass of the same body on the moon surface is (A) \(\mathrm{M} / 6\) (B) 56 (C) \(\mathrm{M}\) (D) None of these

A geostationary satellite is orbiting the earth at a height of \(5 \mathrm{R}\) above that of surface of the earth. \(\mathrm{R}\) being the radius of the earth. The time period of another satellite in hours at a height of $2 \mathrm{R}\( from the surface of earth is \)\ldots \ldots \ldots .$ hr (A) 5 (B) 10 (C) \(6 \sqrt{2}\) (D) \(6 / \sqrt{2}\)

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
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