Planets A and B: Set Up Inequality

[harpazo]

According to Newton’s Law of universal gravitation, the attractive force F between two bodies is given by

F = G m•M/r^2.

Note:
m, and M are the masses of the two bodies

r = distance between the two bodies

G = gravitational constant 6.6742 × 10^(–11) Newton' meter^2 kilogram^(–2)

Suppose an object is traveling directly from Planet A to Planet B. The mass of Planet A is 17.56 • 10^(10) kilograms, the mass of Planet B is 5.34 •10^(5) kilograms, and the mean distance r from Planet A to Planet B is 20,000 kilometers.

Set up an inequality that would answer this question: For an object between Planet A and Planet B, how far from Planet A is the force on the object due to the force of Planet B greater than the force on the object in terms of Planet A?

What is the best way to begin forming the inequality needed here?

 

[Otis]

... What is the best way to begin forming the inequality needed here?

Pick a symbol to represent the unknown quantity asked for and then state its definition.

?

 

[MarkFL]

I would let:

\(m_O\) = the mass of the object

\(m_A\) = the mass of planet A

\(m_B\) = the mass of planet B

\(r\) = the distance from the center of mass of planet A to the center of mass of planet B in meters

\(x\) = the distance from the center of mass of planet A to the center of mass of the object

\(F_A\) = the magnitude of the gravitational force exerted by planet A on the object

\(F_B\) = the magnitude of the gravitational force exerted by planet B on the object

And so we may write:

[MATH]F_A=G\frac{m_Om_A}{x^2}[/MATH]
[MATH]F_B=G\frac{m_Om_B}{(r-x)^2}[/MATH]
And we want:

[MATH]F_B>F_A[/MATH]
Can you proceed?

 

[harpazo]

I would let:

\(m_O\) = the mass of the object

\(m_A\) = the mass of planet A

\(m_B\) = the mass of planet B

\(r\) = the distance from the center of mass of planet A to the center of mass of planet B in meters

\(x\) = the distance from the center of mass of planet A to the center of mass of the object

\(F_A\) = the magnitude of the gravitational force exerted by planet A on the object

\(F_B\) = the magnitude of the gravitational force exerted by planet B on the object

And so we may write:

[MATH]F_A=G\frac{m_Om_A}{x^2}[/MATH]
[MATH]F_B=G\frac{m_Om_B}{(r-x)^2}[/MATH]
And we want:

[MATH]F_B>F_A[/MATH]
Can you proceed?

From this point on it is plug and chug, right?

 

[MarkFL]

From this point on it is plug and chug, right?

Yes, into the inequality, substitute in for the forces, divide through by any common constant factors, and then solve for \(x\) using legal algebraic steps.

 

[MarkFL]

Before you get started, do you think you're going to find the object will be closer to planet A or Planet B, and why?

 

[harpazo]

Before you get started, do you think you're going to find the object will be closer to planet A or Planet B, and why?

I think the gravitational force from Planet B is greater than the force from Planet A. The object will be closer to Planet B.

 

[MarkFL]

I think the gravitational force from Planet B is greater than the force from Planet A. The object will be closer to Planet B.

Planet A is more massive than planet B, and so it will have the greater gravitational force, at some given distance. So, if the object were halfway between the two planets, the force from planet A would be greater, so the object would have to move closer to planet B in order to increase the gravitational force from planet B, while simultaneously decreasing the gravitational force from planet A.

It's always good to think about what we should expect first, so that if we get an answer that seems not to be in line with our expectations, we can check to see if our expectations were wrong, or if our algebra is wrong.

 

[MarkFL]

To follow up, we get the following inequality:

[MATH]G\frac{m_Om_B}{(r-x)^2}>G\frac{m_Om_A}{x^2}[/MATH]
Divided through by [MATH]Gm_O[/MATH] to get:

[MATH]\frac{m_B}{(r-x)^2}>\frac{m_A}{x^2}[/MATH]
Arrange as:

[MATH]\frac{m_B}{(r-x)^2}-\frac{m_A}{x^2}>0[/MATH]
Combine terms on the LHS:

[MATH]\frac{m_Bx^2-m_A(r-x)^2}{(x(r-x))^2}>0[/MATH]
Now, we are only interested in when the object is somewhere in between the planets, or \(0<x<r\), and so our inequality reduces to:

[MATH]m_Bx^2-m_A(r-x)^2>0[/MATH]
Expand:

[MATH]m_Bx^2-m_A(r^2-2rx+x^2)>0[/MATH]
Distribute and arrnge in standard form:

[MATH](m_B-m_A)x^2+2m_Arx-m_Ar^2>0[/MATH]
By the quadratic formula, the critical value is (recall we must discard the negative root):

[MATH]x=\frac{-2m_Ar+\sqrt{4m_A^2r^2+4(m_B-m_A)m_Ar^2}}{2(m_B-m_A)}=\frac{-m_Ar+r\sqrt{m_Bm_A}}{m_B-m_A}=\frac{r\sqrt{m_A}(\sqrt{m_B}-\sqrt{m_A})}{m_B-m_A}=\frac{r}{\sqrt{\dfrac{m_B}{m_A}}+1}[/MATH]
Let's use as our test point:

[MATH]x=\frac{1}{2}\cdot\frac{r}{\sqrt{\dfrac{m_B}{m_A}}+1}[/MATH]
Our inequality becomes:

[MATH]m_B\left(\frac{1}{2}\cdot\frac{r}{\sqrt{\dfrac{m_B}{m_A}}+1}\right)^2-m_A\left(r-\left(\frac{1}{2}\cdot\frac{r}{\sqrt{\dfrac{m_B}{m_A}}+1}\right)\right)^2>0[/MATH]
[MATH]m_B-\left(2\sqrt{m_B}+\sqrt{m_A}\right)^2>0[/MATH]
[MATH]3m_B+4\sqrt{m_Am_B}+m_A<0[/MATH]
This obviously cannot be true, thus our solution is on:

[MATH]\left(\frac{r}{\sqrt{\dfrac{m_B}{m_A}}+1},r\right)[/MATH]
Now, let's return to our critical value, representing the distance the object must travel to reach the point where the gravitational forces cancel one another out:

[MATH]x=\frac{r}{\sqrt{\dfrac{m_B}{m_A}}+1}[/MATH]
Here are some follow up questions, which I hope you will give due consideration and answer.

What happens to this value as the mass of planet A increases without bound, while the mass of planet B remains fixed? And when the mass of planet B increases without bound, while the mass of planet A remains fixed? Do these results agree with intuition?

Suppose we are told the mass of the Earth is roughly 81 times that of the moon. What percentage of the way between the Earth and the Moon is the point where an object would be pulled towards both with equal force, but in opposite directions, that is, where the net gravitational force (from just the Earth and the Moon) on the object is zero?

Can you now apply this formula to your other problem?

 

[harpazo]

To follow up, we get the following inequality:

[MATH]G\frac{m_Om_B}{(r-x)^2}>G\frac{m_Om_A}{x^2}[/MATH]
Divided through by [MATH]Gm_O[/MATH] to get:

[MATH]\frac{m_B}{(r-x)^2}>\frac{m_A}{x^2}[/MATH]
Arrange as:

[MATH]\frac{m_B}{(r-x)^2}-\frac{m_A}{x^2}>0[/MATH]
Combine terms on the LHS:

[MATH]\frac{m_Bx^2-m_A(r-x)^2}{(x(r-x))^2}>0[/MATH]
Now, we are only interested in when the object is somewhere in between the planets, or \(0<x<r\), and so our inequality reduces to:

[MATH]m_Bx^2-m_A(r-x)^2>0[/MATH]
Expand:

[MATH]m_Bx^2-m_A(r^2-2rx+x^2)>0[/MATH]
Distribute and arrnge in standard form:

[MATH](m_B-m_A)x^2+2m_Arx-m_Ar^2>0[/MATH]
By the quadratic formula, the critical value is (recall we must discard the negative root):

[MATH]x=\frac{-2m_Ar+\sqrt{4m_A^2r^2+4(m_B-m_A)m_Ar^2}}{2(m_B-m_A)}=\frac{-m_Ar+r\sqrt{m_Bm_A}}{m_B-m_A}=\frac{r\sqrt{m_A}(\sqrt{m_B}-\sqrt{m_A})}{m_B-m_A}=\frac{r}{\sqrt{\dfrac{m_B}{m_A}}+1}[/MATH]
Let's use as our test point:

[MATH]x=\frac{1}{2}\cdot\frac{r}{\sqrt{\dfrac{m_B}{m_A}}+1}[/MATH]
Our inequality becomes:

[MATH]m_B\left(\frac{1}{2}\cdot\frac{r}{\sqrt{\dfrac{m_B}{m_A}}+1}\right)^2-m_A\left(r-\left(\frac{1}{2}\cdot\frac{r}{\sqrt{\dfrac{m_B}{m_A}}+1}\right)\right)^2>0[/MATH]
[MATH]m_B-\left(2\sqrt{m_B}+\sqrt{m_A}\right)^2>0[/MATH]
[MATH]3m_B+4sqrt{m_Am_B}+m_A<0[/MATH]
This obviously cannot be true, thus our solution is on:

[MATH]\left(\frac{r}{\sqrt{\dfrac{m_B}{m_A}}+1},r\right)[/MATH]
Now, let's return to our critical value, representing the distance the object must travel to reach the point where the gravitational forces cancel one another out:

[MATH]x=\frac{r}{\sqrt{\dfrac{m_B}{m_A}}+1}[/MATH]
Here are some follow up questions, which I hope you will give due consideration and answer.

What happens to this value as the mass of planet A increases without bound, while the mass of planet B remains fixed? And when the mass of planet B increases without bound, while the mass of planet A remains fixed? Do these results agree with intuition?

Suppose we are told the mass of the Earth is roughly 81 times that of the moon. What percentage of the way between the Earth and the Moon is the point where an object would be pulled towards both with equal force, but in opposite directions, that is, where the net gravitational force (from just the Earth and the Moon) on the object is zero?

Can you now apply this formula to your other problem?

Thank you for the follow-up reply. I can apply this great formula to the other problem.

 

[MarkFL]

I will answer your questions but only request help if needed.

I was hoping you would answer them in this thread. At this point though, I have, shall we say, moved on.

 

[MarkFL]

Now, let's return to our critical value, representing the distance the object must travel to reach the point where the gravitational forces cancel one another out:

[MATH]x_C=\frac{r}{\sqrt{\dfrac{m_B}{m_A}}+1}[/MATH]
What happens to this value as the mass of planet A increases without bound, while the mass of planet B remains fixed?

We can see that as \(m_A\) grows without bound the fraction [MATH]\frac{m_B}{m_A}[/MATH] goes to zero, and so \(x_C\) goes to \(r\). This agrees exactly with intuition.

And when the mass of planet B increases without bound, while the mass of planet A remains fixed?

We can see that as \(m_B\) grows without bound the fraction [MATH]\frac{m_B}{m_A}[/MATH] also grows without bound, and so \(x_C\) goes to \(0\). This also agrees exactly with intuition.

This is one reason I worked problems in general terms as a student...so I could look at what happens to the solution as the parameters change in various ways. This leads to a deeper understanding of the problems, and how the solutions behave.

For this formula, I didn't think it unreasonable to think you could "eyeball" these results in a matter of seconds. After all, the only thing you need to keep in mind is how a fraction behaves as either the numerator or denominator grows without bound.

Suppose we are told the mass of the Earth is roughly 81 times that of the moon. What percentage of the way between the Earth and the Moon is the point where an object would be pulled towards both with equal force, but in opposite directions, that is, where the net gravitational force (from just the Earth and the Moon) on the object is zero?

In this case the Earth is planet A and the Moon is planet B. We could write:

[MATH]\frac{x_C}{r}=\frac{\dfrac{r}{\sqrt{\dfrac{m_B}{m_A}}+1}}{r}=\frac{1}{\sqrt{\dfrac{m_B}{81m_B}}+1}=\frac{1}{\dfrac{1}{9}+1}=\frac{9}{10}=90\%[/MATH]
And so we find that we must travel 90% of the way between the center of the Earth and the center of the Moon to reach to point where the net gravity from these two bodies is zero.

I know I said I had moved on, but I hate leaving things undone, and I wanted my questions to be answered for the benefit of other students who may read the thread.