UNIVERSE
SATELLITESICY BODIES
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Question
[CLICK ON ANY CHOICE TO KNOW THE RIGHT ANSWER]
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7.73 km s^-1
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10.9 km s^-1
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36.4 km s^-1
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244 km s^-1
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Detailed explanation-1: -To stay in orbit, a satellite has to travel at a very high velocity, which depends on the height. So, typically, for a circular orbit at a height of 300 km above the Earth’s surface, a speed of 7.8 km/s (28, 000 km/h) is needed. At this speed, the satellite will complete one orbit around the Earth in 90 minutes.
Detailed explanation-2: -On earth, the escape velocity is around 40, 270 kmph, which is around 11, 186 m/s.
Detailed explanation-3: -The escape velocity vesc is expressed as vesc = Square root of√2GMr, where G is the gravitational constant, M is the mass of the attracting mass, and r is the distance from the centre of that mass.
Detailed explanation-4: -The escape velocity is a multiplicative factor greater than the orbital velocity: v = sqrt(2GM/r) (escape velocity = sqrt(2) * orbital velocity) If an object achieves escape velocity, it leaves its orbit and is no longer gravitationally bound to the object it was orbiting.