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What is the escape velocity of a satellite around Venus if its specific angular momentum is 47,862.73 km2/s? The orbit of satellite is circular and gravitational parameter of Venus is 324,859 km3/s2.

What is the escape velocity of a satellite around Venus if its specific angular momentum is 47,862.73 km2/s? The orbit of satellite is circular and gravitational parameter of Venus is 324,859 km3/s2. Correct Answer 9.599 km/s

Given, Specific angular momentum (h) = 47,862.73 km2/s Gravitational parameter (μ) = 324,859 km3/s2 Escape velocity (vesc) = 21/2(μ/h) = 21/2*(324,859/47,862.73) = 9.599 km/s

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S1:The path of Venus lies inside the path of the Earth.
P :When at its farthest from the Earth, Venus is 160 million away
Q :With such a wide range between its greatest and leat distances it is natural that at sometimes Venus appears much brighter than at others.
R :No other body ever comes so near the Earth, with the exception of the Moon and occasional comet or asteroid.
S :When Venus is at its nearest to the earth it is only 26 million miles away.
S6:When at its brightest, it is easily seen with the naked eye in broad daylight.

The Proper sequence should be:
S1: The path of Venus lies inside the path of the Earth.
P: When at its farthest from the Earth, Venus is 160 million miles away.
Q: With such a wide range between its greatest and least distances it is natural that at sometimes Venus appears much brighter than others.
R: No other body ever comes so near the Earth, with the exception of the Moon and an occasional comet or asteroid.
S: When Venus is at its nearest to the Earth, it is only 26 million miles away.
S6: When at its brightest, it is easily seen with the naked aye in broad daylight.

The Proper sequence should be:
Deuteron in its ground state has a total angular momentum J = 1 and a positive parity. The corresponding orbital angular momentum L and spin angular momentum S combinations are