Question

A satellite of mass 220 kg is placed into Earth orbit at a height of 850 km above the surface.

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

(b) What is the satellite's speed?
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.
J

Answer 1

a. Use Newton's Law of Gravitation .

Consider that centripeatal force must be equal to gravitational force.

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

Simplifying the above expression

T = sqrt(4*pi^2*r^3/GM)

= sqrt(4*pi^2*(6.37x10^6+ 850x10^3)^3/(6.67x10^-11*5.98x10^24)

T = 6100 s

T = 1.7 hrs

b. Calculate the satellite speed.

v = sqrt(GM/r)

= sqrt(6.67x10^-11*5.98x10^24/(6.37x10^6+ 0.85x10^6))

= 7432.67 m/s

c. Consider the conservation of energy

(K + U)earth = ( K+U)orbit

here 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*220*(1/(6.37x10^6) - 1/(6.37x10^6 + 0.85x10^3)) + 0.5*220*(7432.67)^2

= 7.7 x10^9J