A spacecraft of 100 kg mass is in a circular orbit about the Earth at a height h

Question

A spacecraft of 100 kg mass is in a circular orbit about the Earth at a height h = 3RE.

(a) What is the period of the spacecraft's orbit about the Earth?

T= ______h

(b) What is the spacecraft's kinetic energy?

K=_______J

(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.)

L=_______

d) Find the numerical value of the angular momentum.

L=______J*s

Explanation / Answer

a) we know time periode of Spacecraft, T = 2*pi*r^(3/2)/sqrt(G*Me)

= 2*pi*(4*Re)^1.5/sqrt(G*Me)

= 2*pi*(4*6.37*10^6)^1.5/sqrt(6.67*10^-11*5.97*10^24)

= 40496 s

= 11.24 hours

b) Orbital speed, Vo = sqrt(G*Me/r)

= sqrt(6.67*10^-11*5.97*10^24/(4*6.37*10^6))

= 3953 m/s

so, kinetic enrgy = 0.5*m*vo^2

= 0.5*100*3953^2

= 7.81*10^8 J

c) Ngular momentumL = m*vo*r

= m*vo*r

= 2*0.5*m*vo^2*r/(vo)

= 2*KE*r/vo

= 2*KE*(4*Re)/sqrt(G*Me/(4*Re))

= 2*KE*(4*Re)*sqrt(4*Re/(G*Me))

d) L = 2*7.81*10^8*(4*6.37*10^6)sqrt(4*6.37*10^6/(6.67*10^-11*5.97*10^24))

= 1*10^13 J.s

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