"Calculating Work Needed to Change Orbits

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To calculate the work needed for a shuttle to deploy a satellite from an orbit at 400 km to 600 km, the correct approach involves using gravitational potential energy (U) and kinetic energy (E_k) equations. Initial calculations yielded inconsistent results, prompting a discussion on measuring distances from the center of mass. The total energy of the satellite is not constant during the transition, leading to confusion between gravitational potential energy and kinetic energy calculations. A simpler formula for small changes in radius was suggested for clarity. The conversation emphasizes the importance of correctly applying gravitational equations to achieve accurate results.
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Homework Statement



A shuttle orbiting the Earth at 400km deploys a satellite of mass 800kg into orbit a further 200km from earth. Calculate the work that must be done by the shuttle to deploy the sattelite.

Homework Equations



E_{k} = \frac{GMm}{2r}, U = -\frac{GMm}{r}

The Attempt at a Solution



I tried using W = \Delta E_{k} to solve the problem:

800GM\left(\frac{1}{2 \times 600000} - \frac{1}{2 \times 400000}\right)=-1.334 \times 10^{11} J

But the solutions gives 1.4 \times 10^9 J

Moreover, using the following method gives a different answer:

\int_{400000}^{600000} \frac{GM(800)}{r^2}dr=2.658 \times 10^{11} J

But I thought E_{tot} = E_{k} + U, so \Delta E_{k} = -\Delta U ??

Could someone tell me why my methods do not work and what the correct method is for dealing with this? Thankyou
 
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a case of mistaken Identity …

Hi Identity! :smile:
Identity said:
A shuttle orbiting the Earth at 400km deploys a satellite of mass 800kg into orbit a further 200km from earth. Calculate the work that must be done by the shuttle to deploy the sattelite.

Could someone tell me why my methods do not work

They do work …

but you need to measure r from … ? :wink:
 
Oh right I need to measure r from the centre of mass XD thanks tiny-tim

But

800GM\left(\frac{1}{2(6.4 \times 10^6 + 600000)} - \frac{1}{2(6.4 \times 10^6 + 400000)}\right)=-6.7 \times 10^8 J

which is still different!

Also, when the satellite moves from 400km to 600km, is the total energy of the satellite constant? If so, why do I get different answers when working with GPE as opposed to Kinetic energy?
 
Identity said:
800GM\left(\frac{1}{2(6.4 \times 10^6 + 600000)} - \frac{1}{2(6.4 \times 10^6 + 400000)}\right)=-6.7 \times 10^8 J

which is still different!

WIhtout seeing the details of your calculation, I can't check it …

but it would be a lot easier to use the formula 1/r - 1/(r + ∆) ~ ∆/r2 :wink:

anyway, going to bed now … :zzz:
 
lol I just plugged the whole calculation into my calculator...

Sorry I'm having some trouble understanding. Can you show me how you would do it?
 
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