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A satellite is given a boost of 550 MJ of energy to move it from its initial orb

ID: 1780813 • Letter: A

Question

A satellite is given a boost of 550 MJ of energy to move it from its initial orbit at an altitude of 150 km to a higher altitude orbit. If the satellite has a mass of 1.07 × 103 kg, what is the new altitude it reaches? Take the mass of the Earth to be 5.97 x 1024 kg and its radius to be 6.371 x 105 m. 205.2 Consider the gravitational force between the satellite and Earth. How is this force related to the centripetal force that keeps the satellite in its circular orbit? What is the difference in energy of the system between the two orbits? km

Explanation / Answer

Kinetic energy in an orbit

The kinetic energy of an object in orbit can easily be found from the following equations:

Centripetal force on a satellite of mass m moving at velocity v in an orbit of radius r = mv²/r

But this is equal to the gravitational force (F) between the planet (mass M) and the satellite (mass m):

F =GMm/r² and so mv² = GMm/r

But kinetic energy = ½mv² and so:

kinetic energy of the satellite = ½ GMm/r

therefore total energy of a satellite in orbit of radius r is

-GMm/r + GMm/2r = -GMm/2r

total energy of satellite at height of 150 km = - GMm/2(R + 150000)

M = 5.97 x10^24 kg

R = 6.371 x10^6 m

m = 1.07x10^3 kg

G = 6.67×1011 N·(m/kg)2

if new altitude is x m

from conservation of energy

=> - GMm/2(R + 150000) + 550 x 10^6 = - GMm/2(R + x)

=> 550 x 10^6 = (GMm/2) [1/(R + 150000) - 1/(R + x)]

=> 550 x 10^6 = (GMm/2) [R + x - (R + 150000)]/[(R + x)(R + 150000)]

=> 550 x 10^6 = (GMm/2) [x - 150000]/[(R + x)(R + 150000)]

=> (550 x 10^6)*2 = (6.67×10^11)(5.97 x10^24)(1.07x10^3 ) [x - 150000]/[(R + x)(R + 150000)]

=> (12.9*2)[(R + x)(R + 150000)] = 10^10[x - 150000]

=> 25.8[R² + 150000R + Rx + 150000x] = 10^10(x) - (1.5)10^15]

=> 25.8[R² + 150000R] + 1.5(10^15) = x[10^10 -25.8R – 25.8(150000)]

=> 2.572 * 10^15 = x (9.831758*10^9)

=> x = 261,587.85 m = 262 km

Ans: the new altitude it reaches is ~= 262 km

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