A spacecraft of 130 kg mass is in a circular orbit about the Earth at a height h

Question

A spacecraft of 130 kg mass is in a circular orbit about the Earth at a height h 4RE (a) What is the period of the spacecraft's orbit about the Earth? (b) What is the spacecraft's kinetic energy? (c) Express the angular momentum L of the spacecraft about the center of the Earth in terms of its kinetic energy K. (Use the following as necessary: RE for the radius of the Earth, K for the kinetic energy of the satellite, and m for the mass of the satellite.) (d) Find the numerical value of the angular momentum

Explanation / Answer

a) we know, Re = 6.37*10^6 m

h = 4*Re

radius of orbit of space craft, r = Re + h

= Re + 4*Re

= 5*Re

= 5*6.37*10^6 m

= 3.185*10^7 m

we know, T = 2*pi*r^(3/2)/sqrt(G*Me)

= 2*pi*(3.185*10^7)^1.5/sqrt(6.67*10^-11*5.98*10^24)

= 56550 s

= 56550/(60*60)

= 15.7 hours

b) orbital speed, vo = sqrt(G*Me/r)

= sqrt(6.67*10^-11*5.98*10^24/(3.185*10^7))

= 3538 m/s

KE = (1/2)*m*v^2

= (1/2)*130*3538^2

= 8.14*10^8 J

c) L = m*v*r

= p*5*Re

= sqrt(2*m*K)*5*Re

d) L = m*v*r

= 130*3538*3.185*10^7

= 1.46*10^13 J.s

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