A satellite of mass 210 kg is placed into Earth orbit at a height of 500 km abov

Question

A satellite of mass 210 kg is placed into Earth orbit at a height of 500 km above the surface. Assuming a circular orbit, how long does the satellite take to complete one orbit? It may be easiest to answer part (find the speed of the satellite) first. h What is the satellite's speed? If you know the net force acting on an object that moves in a circle, how can you calculate its speed? m/s 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. The satellite has some initial kinetic energy on the surface of the Earth because of the Earth's rotation. What is the total energy of the satellite when it is orbit? J

Explanation / Answer

1. Use Newton's Law of Gravitation in the form (You have to use this form because g is not constant)

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+ 500x10^3)^3/(6.67x10^-11*5.98x10^24) = 5662s

= 5662/3600 = 1.57hrs

2. v = sqrt(GM/r) = sqrt(6.67x10^-11*5.98x10^24/(6.37x10^6+ 0.5x10^6)) = 7619m/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 + 0.5x10^3)) + 1/2*210*(7619)^2

= 7.05x10^9J

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