If ve and vo represent the escape velocity and orbital velocity of a satellite corresponding to a circular orbit of radius R respectively, then
(A) ve = vo
(B) \(\sqrt{2}\)vo = ve
(C) ve = \(\frac{1}{\sqrt{2}}\)vo
(D) ve and vo are not related
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A circular coil `P` of `100` turns and radius `2cm` is placed coaxially at the center of another circular coil `Q` of `100` turns and radius `20cm` .
Correct Answer - `(a) 3.94xx10^(-4)H (b) 19.72xx10^(-3)V (c ) 19.72xx10^(-3)Wb//s (d) 9.86xx10^(-5)C`
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Assuming that a planet goes round the sun in a circular orbit of radius `0.5AU` determine the angle of maximum elongation for the planet and its dista
Correct Answer - `30^(@), 0.866AU`
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The binding energy of a satellite revolving around planet in a circular orbit is 3 × 10^9 J. Its kinetic energy is ______
Correct answer is (D) 3 × 10+9 J
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Define escape velocity of a satellite.
The minimum velocity with which a both’ should he thrown vertically upwards from the surface of the Earth so that it escapes the Earth ‘s gravitational field, is called the...
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Write the answer of the following question in one word- (a) What is the orbital speed of Geo-stationary satellite ? (b) For a satellite moving in an o
Correct Answer - (a) 3.1 km/s; (b) `-1/2`
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A satellite is orbiting around a planet at a constant height in a circular orbit. If the mass of the planet is reduced to half, the satellite would
Correct option is: (D) escape from the planet.
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Escape velocity on a planet is ve. If radius of the planet remains same and mass becomes 4 times, the escape velocity becomes
Correct option is: (B) \(2v_e\) Escape Velocity \(v_e = \sqrt {2gRe}\) \(\frac{V_{e1}}{V_{e2}} \propto \sqrt \frac{4R_1}{R_1}\) \(V_{escape \ 2} = 2 \ V_{escape \ 1}\)
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Escape velocity on a planet is ve. If radius of the planet remains same and mass becomes 4 times, the escape velocity becomes
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If the escape velocity of a body on Earth is 11.2 km/s, the escape velocity of the body thrown at an angle 45° with the horizontal will be
Correct option is: (A) 11.2 km/s Escape velocity does not depend on the angle of projection. \(V_{escape} = \sqrt {2gRe} \)
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