Calculate Energy of Satellite Orbiting Earth at r = 13/2R | Homework Solution

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The discussion focuses on calculating the energy required to launch a satellite into a circular orbit at a radius of r = 13/2R from Earth. Key parameters include the mass of the satellite (6000 kg), the mass of Earth (5.98 x 10^24 kg), and the gravitational constant (G = 6.67259 x 10^-11 Nm²/kg²). The user initially calculated the total energy as -2.8911 x 10^10 J but found it to be incorrect, leading to confusion about the proper formula to use. They realized that the distance for work should account for the height above Earth's surface, adjusting the calculation accordingly. Ultimately, the user successfully found the correct answer.
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Homework Statement


msatellite = 6000 kg ;
Mearth = 5:98 £ 1024 kg ;
Rearth = 6:37 £ 106 m; and
G = 6:67259 £ 10¡11 Nm2=kg2 :
The radius of the satellite around the center
of the Earth is r =13/2R.
Hint: You may Find it useful to take into account that the gravitational force is a conservative force.
Hint: The universal gravitational force law
is F= GMm/r^2.

Caution: Neglect the rotational kinetic energy due to the Earth's rotation.
Find the energy required to launch a satellite from Earth into the circular orbit at the
specified radius r =13/2R. Answer in units of J.

Homework Equations


F = GMm/r^2.

U(gravitational) = GMm/r
kinetic energy (translational) = 1/2mV^2 = (1/2GMm)/r



The Attempt at a Solution



Total energy = Kinetic + Ug = 0.5GMm/r - GMm/r = -2.8911e10 J
but it is wrong
please help
 
Physics news on Phys.org
maybe this will help as well

physics.jpg
 
i realized that for work = F*D, D is not r but r-R,earth because its launched from the surface of the earth, but I'm still getting a wrong answer
please guys, i really could use some help
 
okay, total energy should be K - Ug = .5GMm/r - GMm/r^2 but my answer is still wrong
(i'm getting 2.8911e10)
 
nvr mind, got the answer
 

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