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GDss · Feb 1, 2010, 05:34 AM

hi everyone.
my question is based on the physics topic :gravity
I have a Mass of planet Mars 6.42*10^23 and its radius R 3.40*10^6M

WHAT IS THE MINIMUM ENERGY NEEDED TO LAUNCH THE 230KG SATELLITE FROM THE SURFACE OF THE PLANET?

I deduced the formula
force of centripetal=force of gravity

mv^2/R=mg

v^2=rg

is this formula correct to calculate the answer?

ebaines · Feb 1, 2010, 08:16 AM

Not quite right. Remember that "g" is the acceleration due to gravity on the earth's surface, so does not apply on Mars. You want to use:

mv^2/R = GmM/R^2

where R = radius of orbit, m is the satellite's mass, and M is the mass of Mars.

This will get you the velocity needed for a circular orbit of radius R. However, the energy needed to reach that orbit is the sum of the kinetic energy of that velocity PLUS the potential energy you need to gain to reach that height. The gain in PE with altitude is: GMm/R - GMm/R_m, where R_m is the radius of Mars.

One last thing - in calculating the gain in Kinetic energy needed don't forget that you have a starting velocity due to the rotation of the planet. So if you launch in the direction of rotation you need less additional kinetic energy than if you launch in a different direction. This is a main reason why the Kennedy Space Center is located on the east coast of Florida - you can launch to the east (in the direction of earth's rotation) and immediately be over water for safety reasons, and it's about as close to the equator as you can get in the continental US, thus reducing the amount of additional KE needed to reach orbital velocity.

GDss · Feb 1, 2010, 08:27 AM

Thank you very much!!

GDss · Feb 1, 2010, 08:30 AM

owh yes I just realized. Can't I use the work formula. As the question asks for the min enrgy?
I used W=f*d
where f=G*Mm/R^2
and d=radius
?

ebaines · Feb 1, 2010, 08:51 AM

Yes, you can use work principles. Since the gravitational force varies with distance, you have to calculate the work as:

$<br /> W = \displaystyle \int _{R_M} ^R \frac {Gm M} {r^2} dr = \frac {GMm} {R_M} - \frac {GMm} {R}<br />$

This is the formula I gave you for the work needed to overcome gravity, although I now realize that I had the terms reversed (gotta watch those signs!).

As I read the problem statement it seems that there is nothing about the satellite going into orbit - only about it being "launched." Is there other information given, such as how high does the satellite go? Or perhaps they are looking for the escape velocity from Mars (where effectively $R = \infty$ and $v = 0$ )?

GDss · Feb 1, 2010, 09:12 AM

The question asked the minimum energy required to launch the satellite from the surface to the orbit at a certain altitude equal to R. which is the orbit. I assume that if they asked for the minimum energy then the answer would be in joules. Thus I don't think they asked for escape velocity.
THANKS SO MUCH ON THE FORMULA