Question

A satellite of mass 210 kg is placed into Earth orbit at a height of 450 km above the surface.

(a) Assuming a circular orbit, how long does the satellite take
to complete one orbit?

1.56 h

(b) What is the satellite's speed?

7647.53 m/s

(c) Starting from the satellite on the Earth's surface, what is the
minimum energy input necessary to place this satellite in orbit?
Ignore air resistance but include the effect of the planet's daily
rotation.

Answer #1

GMm/r^2 = m*v^2/r = m*4?^2*r/T^2

so T = sqrt(4?^2*r^3/GM) = sqrt(4?^2*(6.37x10^6+
850x10^3)^3/(6.67x10^-11*5.98x10^24) = 6103s

= 6103/3600 = 1.695hrs

2. v = sqrt(GM/r) = sqrt(6.67x10^-11*5.98x10^24/(6.37x10^6+
850x10^3)) = 7423m/s

3. Use conservation of energy here

K + U)earth = K+U)orbit where U = -GMm/r

So K = U orbit - U earth + K orbit

= - GMm/r -(-GMm/R) + 1/2*m*v^2

= GMm*(1/R - 1/r) + 1/2*m*v^2

= 6.67x10^-11*5.98x10^24*210*(1/(6.37x10^6... - 1/(6.37x10^6 +
850x10^3)) + 1/2*210*(7423)^2

= 7.35x10^9J

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