Physics, asked by hemalathagogula1986, 1 year ago

A satellite of mass 600kg in a circular orbit is at a height of 2000m above the earth's surface. Taking the radius of the Earth to be 6400km and the value of g° to be 1 N/kg. The satellite's kinetic energy is

Answers

Answered by paulaiskander2
3

Answer:

1.87*10^{10}J

Step by step explanation:

It is given that:

  • Mass of satellite = 600 Kg
  • Height of orbit = 2000m
  • Radius of the earth = 6400 km = 6.4*10^6m

Therefore, the distance between the satellite and the center of the earth 'r' = 2000+6.4*10^6=6.402*10^6\:m

Velocity of the satellite = \sqrt{\frac{GM}{r}}; where G is a constant (6.673*10^{-11}), and M is the mass of the Earth (6*10^{24}\:kg).

v=\sqrt{\frac{6.673*10^{-11}*6*10^{24}}{6.402*10^6}}=7908.2\:m/s

K.E.=\frac{1}{2}mv^2=\frac{1}{2}*600*(7908.2)^2=1.87*10^{10}\:J



hemalathagogula1986: Correct answer
Answered by srivan123
0

Answer:

1.92×10^9 J

Explanation:

hope it helps you

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