Calculating Energy Requirements for Placing a Space Shuttle in Orbit

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Homework Help Overview

The discussion revolves around calculating the energy requirements for placing a space shuttle in orbit at an altitude of 1000 km. The subject area includes gravitational potential energy and kinetic energy in the context of orbital mechanics.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the energy needed to overcome gravitational potential energy and to achieve the necessary kinetic energy for orbit. Questions arise regarding the inclusion of the shuttle's initial kinetic energy due to Earth's rotation and the implications of this on the overall energy calculation.

Discussion Status

There is an ongoing examination of the energy equations presented, with participants providing feedback on specific terms and their correctness. Some participants are clarifying the distinction between final and initial kinetic energy, while others are questioning the relevance of Earth's rotational energy in the calculations.

Contextual Notes

Participants note that the initial velocity of the shuttle is influenced by its launch latitude, and there is a mention of the relatively small contribution of Earth's rotation to the shuttle's energy requirements.

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.

[tex]\Delta E_p=-G\frac{Mm}{R+h} + G\frac{Mm}{R}[/tex]

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

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

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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