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One problem that I'm having trouble with (as opposed to the other):

The Messenger is a probe that orbits Mercury $700 \rm km$ from the surface. What is the tangential velocity it should be rotating at so that it doesn't precipitate towards the planet, in $\rm m/s$?

Data: Mercury's mass $3,3 \times 10 ^{23} \rm kg$ Diameter: $4870 \rm km $ . Gravitational constant: $G=6,67 \times 10^{-11} \rm m^{3}kg^{-1}s^{-2}$

I assume I need to use the following $$a_{c}=\frac {V^2} r$$

$$F=G\frac{m_1 m_2}{d^2}$$

EDIT:

Solving for $V$ in $$G\frac{{{m_1}{m_2}}}{{{d^2}}} = {m_1}\frac{{{V^2}}}{d}$$ did it.

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Peter, you appear to be going about this in the right way (conservation of energy plus the acceleration due to circular motion), but we don't work particular problem on this site. That included checking your work. – dmckee♦ Jun 22 '12 at 1:55
Voting to close too specific. If you can edit your question to reflect a specific concept you are having trouble with someone might be able to help. – Argus Jun 22 '12 at 1:55
@dmckee OK. I didn't pay attention to that, my mistake, it is true, this is too specific. The thing is I am going through some mock exams since I have a mid term coming soon. I wanted to show my work, but maybe I can get a second opinion. Thanks anyways. – Peter Tamaroff Jun 22 '12 at 2:23
@PeterT.off: thanks for taking it well! And also, we do appreciate that you showed your work rather than asking for an answer, even though on this site we don't handle these "work-checking" questions. (We're a little more restrictive than other physics Q&A sites in that respect.) I'll mention that I agree with dmckee that your method seems to be quite good. – David Zaslavsky♦ Jun 22 '12 at 2:31
@DavidZaslavsky Oh, I think that's OK. I'm an active member of math.SE so I understand. – Peter Tamaroff Jun 22 '12 at 2:32
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1 Answer

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The answers to your two questions both come out of the observation that you can not equate force to acceleration. To get them into a form where you can equate them alter the second one to get the acceleration of the satellite due to the planets gravity.

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Should I be using Newton's Second Law? – Peter Tamaroff Jun 22 '12 at 2:57
Do you know another way? This kind of observation is one of the more useful tools in solving physics problems---especial those asked in school. You have one acceleration and once force, and you say "Aha! Newton had something to say about that. What happens if I ..." With luck it all works out. What you are looking for here is a relationship in which the mass of the orbiter (which you've noted you do not have) does not appear. – dmckee♦ Jun 22 '12 at 3:01
Right. by the way, this is from an introductory course from an university (a basic common course for the exact sciences and engineering factulty.). – Peter Tamaroff Jun 22 '12 at 3:04
I got it. Thanks. – Peter Tamaroff Jun 22 '12 at 3:21

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