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Physics 9 Online
OpenStudy (anonymous):

. A satellite of mass 1000 kg moves in a circular orbit of radius 7000 km round the earth, assumed to be a sphere of radius 6400 km. Calculate the total energy needed to place the satellite in orbit from the earth.

OpenStudy (anonymous):

think about what energy changes take place between the initial and final states of the satellite

OpenStudy (anonymous):

dont get it ?

OpenStudy (anonymous):

what forms of energy of the satellite do you need to consider ?

OpenStudy (anonymous):

kinetic and gravitational potential energy

OpenStudy (anonymous):

right and you know the formula for kinetic energy ?

OpenStudy (anonymous):

1/2gMm/r

OpenStudy (anonymous):

yes that's right, it's easier than i thought and you need the formula for potential energy as well

OpenStudy (anonymous):

?

OpenStudy (vincent-lyon.fr):

You need to know where you are launching the satellite from. (Pole or Equator won't result in the same answer)

OpenStudy (anonymous):

how so vincent ? initial ke = 0 final ke = GMm/2Rfinal initial pe = -GMm/Rearth final pe = -GMm/Rfinal

OpenStudy (vincent-lyon.fr):

Initial KE is zero if you launch the satellite from the Poles only. If you launch it from the Equator, you can save a lot of fuel because KE is not zero since Earth is rotating, so the satellite's initial velocity is not zero. This is why all launchpads are as close to the Equator as possible: South Florida in the US, Kazakhstan in the former USSR, French Guyana for France.

OpenStudy (anonymous):

I see, good point - does it really make a big difference ? On my simple picture I have an energy of 34 Megajoules needed.

OpenStudy (anonymous):

I suppose really it would also depend on the precise orientation of orbit that was desired, I guess I implicitly assumed that there was no need to worry about such details.

OpenStudy (anonymous):

somewhere i lost a factor of 1000, the mass of the satellite, so the energy should be 34000megajoules i think

OpenStudy (anonymous):

i think the equatorial kinetic energy would contribute about one third of a percent to the required energy at best

OpenStudy (vincent-lyon.fr):

KE energy at the equator is about 108 MJ. It is only a fraction of a %, but you still have to pay for it :-) I am not saying the value is very important. What I am saying is that the question is not consistent. If you work with KE and gravitational PE, you have to train the students to think in the right frame of reference, and this question is misguiding.

OpenStudy (anonymous):

Yes, you are right, it's good practice to know what is being ignored in a simple model. By the way, i suppose the moon is spinning - but I don't like it : )

OpenStudy (vincent-lyon.fr):

Lol

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