Question

A satellite of mass 180 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?

ans)_____h

(b) What is the satellite's speed?

7558.5 Correct: Your answer is correct. 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.

ans)________J

Answer 1

Suppose,

M = Mass of the Earth = 5.98 x 10^24 kg

m = mass of the satellite = 180 kg

R = Radius of the earth = 6.37 x 10^6 m

h = height of the satellite above the Earth = 850 km = 850 x 10^3 m

So, altitude of the satellite from the center of the Earth = r = R+h = (6.37 x 10^6 + 850 x 10^3) m

(a) Use Newton's Law of Gravitation -

GMm/r^2 = m*v^2/r = m*4π^2*r/T^2

=> T = sqrt(4π^2*r^3/GM)

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

= 6103s

Convert this time in hours.

T = 6103/3600 = 1.695 hrs (Answer)

(b) Satellite speed,

v = sqrt(GM/r)

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

= sqrt[(39.89 x 10^13) / (7.22 x 10^6)]

= sqrt(5.525 x 10^7)

= 7433 m/s (Answer)

(c) 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

= 7.35 x 10^9 J (Answer)