the kinetic energy of satellite isK, then its time period of revolutionT is proportiona
Question
the kinetic energy of satellite isK, then its time period of revolutionT is proportiona
Solution
It seems like your question got cut off. However, I can provide a general answer based on what you've asked so far.
The kinetic energy (K) of a satellite in orbit is given by the equation K = 1/2 mv², where m is the mass of the satellite and v is its velocity.
The time period (T) of revolution of a satellite is the time it takes for the satellite to complete one full orbit. This is given by the equation T = 2πr/v, where r is the radius of the orbit.
From these two equations, we can see that the kinetic energy and the time period of revolution are related through the velocity of the satellite. However, they are not directly proportional because the kinetic energy is proportional to the square of the velocity, while the time period is inversely proportional to the velocity.
Therefore, if the kinetic energy of the satellite increases (meaning the satellite is moving faster), the time period of revolution will decrease (meaning it takes less time to complete an orbit). Conversely, if the kinetic energy decreases (meaning the satellite is moving slower), the time period of revolution will increase (meaning it takes more time to complete an orbit).
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