Calculating Energy Requirements for Placing a Space Shuttle in Orbit

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To calculate the energy required to place a 68,000 kg space shuttle in orbit at 1,000 km altitude, both potential and kinetic energy must be considered. The potential energy changes due to the shuttle's altitude, while kinetic energy is influenced by the shuttle's velocity, which is affected by Earth's rotation. The initial kinetic energy from Earth's rotation contributes minimally to the total energy requirement, approximately less than 2%. The distinction between final and initial kinetic energy is crucial, as the shuttle's launch velocity varies with latitude. Understanding these energy components is essential for accurate calculations in orbital mechanics.
Gyroscope

Homework Statement


A space shuttle of 68000 kg is placed in orbit at an altitude of 1000 km from the surface of the Earth. What energy must we provide?

The Attempt at a Solution



I will need to provide energy in order to decrease the potential energy and an additional energy for the kinetic energy of the rotation.

\Delta E_p=-G\frac{Mm}{R+h} + G\frac{Mm}{R}

E_c=\frac{1}{2}mv^2=\frac{1}{2}mGM/(R+h)

So,
E=-G\frac{Mm}{R+h}+ G\frac{Mm}{R}+\frac{1}{2}mGM/(R+h)

Am I right?
 
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You are almost correct, but you have made one 'small' mistake in the first line which has carried through.
 
Thank you for answering, Hootenanny!:rolleyes:

I have just corrected my mistake. My doubt was whether to consider or not the kinetic energy of rotation. Because when the space shuttle reaches an height of 1000 km relative to the surface of the Earth it will need an extra impulse from the fuel engines.
 
The potential energy equation looks fine except you do not square the distance and you have the correct overall idea. The kinetic energy term looks fine as you have used circular motion to derive the velocity.

EDIT: Beaten by Hoot and the coffee machine again!
 
Kurdt, thank you anyway for your kind help!
 
Gyroscope:

The Shuttle has kinetic energy prior to launch. You have solved for the final kinetic energy, not the change in kinetic energy. The distinction is important: It is why the Shuttle is launched to the East.
 
D_H thanks for your reply!

Could you be more specific, please? The kinetic energy prior to launch of the shuttle is the energy associated with the rotation motion of the Earth?
 
Gyroscope said:
D_H thanks for your reply!

Could you be more specific, please? The kinetic energy prior to launch of the shuttle is the energy associated with the rotation motion of the Earth?

Yes, it is.

Although I'm wondering if you're supposed to worry about that for this problem. Your initial velocity will depend on your latitude. At the equator, the launch site is moving 465 m/sec. Your speed anywhere else is equal to the cosine of the latitude times 465. Even at the equator, the rotation of the Earth will contribute less than 2% of the shuttle's energy.
 
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