Question about circular motion of planets

[RoboNerd]

Homework Statement

Homework Equations

I know that Fc = m * centripetal acceleration, where the Fc = the force of gravity

The Attempt at a Solution

[/B]
I was able to use process of elimination to get B for 25 correctly, but I am not able to understand how it is the right answer.

How can I find that the right answers for 25 is B and 26 A. I tried solving with different methods, but I keep hitting the wall.

The method of solving must have something to do with the gravitational force between the planet and satellite.

Thanks in advance

 

[CWatters]

I haven't checked that the answers you mention are correct but try writing equations for both satellites. And then substitute one into the other to eliminate unknowns

 

[RoboNerd]

I already tried that, and I got absolutely nowhere after a large amount of struggle.
Physicsforums here is my last resort for help as I am desperate.

Thanks for the help.

 

[gneill]

For #25 you might look into a series expansion of the result, using the hint that r >> h.

You should know an expression for centripetal force based on mass, radius, and velocity. You should also know an expression for the velocity of a circular orbit given the mass of the planet and the radius of the orbit. If you form ratios lots of unnecessary things should cancel. Call the first orbit "a" and the second "b". Write $F_b/F_a$ and simplify.

 

[RoboNerd]

Hi, I took your advice for doing the ratio, and did the following:

Fa G * mass planet * mass satellite / (h+r)^2
------ = ----------------------------------------------------
Fb G * mass planet * mass satellite / (h+2r)^2

My ratios cancel:

Fa 1/ (h+2r)^2
------ = ----------------------------------------------------
Fb 1/ (h+r)^2

I got Fb = Fa * [ (h+r)^2 / (h+2r)^2 ].

How can I continue on and infer that B is the right answer for 25? And how would I then move from there and solve 26?

Thanks for the help.

 

[haruspex]

(h+2r)

Did you mean that?

When you have corrected that, use the fact that r>>h. That allows you to use the first order approximation for the binomial theorem: if x is small then (1+x)ⁿ is approximately 1+xn.

 

[RoboNerd]

No, sorry. I meant: r + 2h.

This question comes from an ap physics c mechanics practice test, which is for high schoolers and not in-depth about advanced topics, such as the binomial theorem.
I personally know some, but I have forgotten the most of it.

Is there any way to solve this without binomial theorem, and other similarly advanced methods?

Thanks

 

[haruspex]

No, sorry. I meant: r + 2h.

This question comes from an ap physics c mechanics practice test, which is for high schoolers and not in-depth about advanced topics, such as the binomial theorem.
I personally know some, but I have forgotten the most of it.

Is there any way to solve this without binomial theorem, and other similarly advanced methods?

Thanks

As you can see from the question statement, it involves making an approximation based on h<<r. You can either use the binomial theorem or the equivalent expressed in terms of differentials: $\Delta f(x) \approx f'(x)\Delta x$.

 

[RedDelicious]

No, sorry. I meant: r + 2h.

This question comes from an ap physics c mechanics practice test, which is for high schoolers and not in-depth about advanced topics, such as the binomial theorem.
I personally know some, but I have forgotten the most of it.

Is there any way to solve this without binomial theorem, and other similarly advanced methods?

Thanks

You should definitely be comfortable with using the binomial theorem for approximations.

But, as a future strategy, given that this was multiple choice, you could have also determined the correct answer by performing a quick polynomial division - which is taught in a standard high school algebra II class.

As you found above

\frac{F_b}{F_a}=\frac{\left(r+h\right)^2}{\left(r+2h\right)^2}=1-\frac{\left(2rh+3h^2\right)}{r^2+4hr+4h^2}

Which is the only answer that has the factor in the correct form of 1 - u

You can actually take it a step further by expanding, dividing out by h^2, and then making some "common sense" reductions based on r >> h, and you'll arrive at the exact answer. But this is best used when you're anticipating a certain answer and just want to make sure you can actually get it, but in this case it wasn't necessary.

 

[RoboNerd]

Thanks for that approach. Interesting to think about it that way.

Hmm... this multiple choice is intended to be solved in a single minute. This approach with divisions takes a long time... and it is intended for high schoolers who know nothing about the binomial theorem, just basic calculus with integration and differentiation. I seriously believe that the test authors did not want people to solving it with such an involved method. Are there not more simple methods of solving this one, or is this the only way?

Also, could we please take a look at number 26?

Thanks for the help.

 

[haruspex]

with integration and differentiation

My post #8 shows how to do it with differentiation.
(This is really more advanced than using the binomial theorem, in the sense that standard derivatives are obtained by using that theorem. But you may have been taught one and not the other.)

 

[RedDelicious]

Thanks for that approach. Interesting to think about it that way.

Hmm... this multiple choice is intended to be solved in a single minute. This approach with divisions takes a long time... and it is intended for high schoolers who know nothing about the binomial theorem, just basic calculus with integration and differentiation. I seriously believe that the test authors did not want people to solving it with such an involved method. Are there not more simple methods of solving this one, or is this the only way?

Also, could we please take a look at number 26?

Thanks for the help.

I wasn't thinking about it, but even more simply you could have done

\frac{\left(r+h\right)^2}{\left(r+2h\right)^2}=\left(\frac{r+h}{r+2h}\right)^2=\left(1-\frac{h}{r+2h}\right)^2\approx \left(1-\frac{h}{r}\right)^2

and that only takes a matter of seconds to do.

As for number 26,

You know the centripetal force acting on each. You can relate that force to the kinetic energy, and then you just relate the kinetic energies to each other. I'll give you a hint.

\frac{F_a\left(r+h\right)}{2}=

 

[RoboNerd]

OK here's what I did.

I solved for Force of gravity to be equal to = G * m1 * m2 / r^2.

Then I said that Fg = mv^2 / r for centripetal accelerating objects.

So then I solved to get that K.E. = (1 / 2) Fg * r.

Right so far?

Fa(r+h)2=

Is that supposed to be from a ratio of the kinetic energies?

 

[RedDelicious]

OK here's what I did.

I solved for Force of gravity to be equal to = G * m1 * m2 / r^2.

Then I said that Fg = mv^2 / r for centripetal accelerating objects.

So then I solved to get that K.E. = (1 / 2) Fg * r.

Right so far?
Is that supposed to be from a ratio of the kinetic energies?

You have the right idea, but what is r?

 

[Chestermiller]

Consider $\frac{1+x}{1+2x}$.

From geometric progression experience, $\frac{1}{1+2x}=1-2x+(2x)^2...$. So,

$$\frac{1+x}{1+2x}=(1+x)(1-2x+(2x)^2...)=1-x+2x^2...$$

 

[RoboNerd]

You have the right idea, but what is r?

R is an a arbitrary radius, where I can substitute "r + 2h" or "r + h" depending on particular problem context

 

[RoboNerd]

Consider 1+x1+2x1+x1+2x\frac{1+x}{1+2x}.

From geometric progression experience, 11+2x=1−2x+(2x)2...11+2x=1−2x+(2x)2...\frac{1}{1+2x}=1-2x+(2x)^2... So,

1+x1+2x=(1+x)(1−2x+(2x)2...)=1−x+2x2...​

Hi. I do not understand what you are doing, nor how this concept can be possibly applied to my physics question. Thanks for the patience and help.

 

[Chestermiller]

Hi. I do not understand what you are doing, nor how this concept can be possibly applied to my physics question. Thanks for the patience and help.

In my post, x = h/r. You are trying to determine an approximation for $\frac{(r+h)^2}{(r+2h)^2}$. If I divide numerator and denominator by r², I get:$$\frac{(r+h)^2}{(r+2h)^2}=\frac{(1+\frac{h}{r})^2}{(1+2\frac{h}{r})^2}=\frac{(1+x)^2}{(1+2x)^2}=\left(\frac{1+x}{1+2x}\right)^2$$
Do you see the connection now to what I did in post #15?

 

[RedDelicious]

R is an a arbitrary radius, where I can substitute "r + 2h" or "r + h" depending on particular problem context

Ok, you had used "r" so I was just making sure you didn't think it was the, r, as in the radius of the planet. Do you see how to solve part B now?

 

[Chestermiller]

Ok, you had used "r" so I was just making sure you didn't think it was the, r, as in the radius of the planet. Do you see how to solve part B now?

In this problem, r is the radius of the planet, r+h is the distance of satellite #1 from the center of the planet, and r+2h is the distance of satellite #2 from the center of the planet. What did you think?

 

[RedDelicious]

In this problem, r is the radius of the planet, r+h is the distance of satellite #1 from the center of the planet, and r+2h is the distance of satellite #2 from the center of the planet. What did you think?

I should have been more clear, but my reply was partly in reference to one of his earlier ones, where the OP told me that F = mv^2/r and since he hadn't arrived at the correct answer, I wanted to make sure he was speaking in general and wasn't mistakingly using that as the force on any of the masses.

 

[Chestermiller]

I should have been more clear, but my reply was partly in reference to one of his earlier ones, where the OP told me that F = mv^2/r and since he hadn't arrived at the correct answer, I wanted to make sure he was speaking in general and wasn't mistakingly using that as the force on any of the masses.

Ah. OK.

 

[RoboNerd]

In my post, x = h/r.

Hi, how do you get this relationship, and what is the meaning of the variable "x". Maybe I lost it in the long continuation of posts on the thread.

 

[RoboNerd]

I should have been more clear, but my reply was partly in reference to one of his earlier ones, where the OP told me that F = mv^2/r and since he hadn't arrived at the correct answer, I wanted to make sure he was speaking in general and wasn't mistakingly using that as the force on any of the masses.

Hi, I am curious as to how I could proceed. I think we cleared up the case about my arbitrary usage of "r" up, right?

 

[Chestermiller]

Hi, how do you get this relationship, and what is the meaning of the variable "x". Maybe I lost it in the long continuation of posts on the thread.

If you are interested in a physical interpretation for x, it is simply the "dimensionless altitude" of the inner satellite. It often helps to work in terms of dimensionless variables, since this is the way that real physical systems behave. In my analysis, x is simply defined x = r/h to simplify the math. Do you see the mathematical advantage? It reduces the number of variables you are working with from 2 to 1.

 

[RoboNerd]

Why do you define x = r/h?

Thanks

 

[Chestermiller]

Why do you define x = r/h?

Thanks

As I already said, It reduces the number of variables you have to deal with from 2 to 1.

 

[RoboNerd]

OK, I see the reason why you definded x = r/h. How would I proceed further?
Thanks!

 

[Chestermiller]

OK, I see the reason why you definded x = r/h. How would I proceed further?
Thanks!

See my post #15.

 

[RoboNerd]

Ah. OK. I see now. Thanks everyone for your inputs!