PE and Torque (EK in-class exam 3)

[Jwinsler7]

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Q1. A rocket is launched from earth to explore our solar system and beyond. As the rocket moves out of the earth's atmosphere and into deep space, the gravitational constant g decreases, and approaches zero, and the gravitational potential energy of the rocket:

A. also decreases and approaches zero
B. continually increases
C. remains constant
D. increases at first and then decreases and approaches zero.

I answered A just from using U=mgh, as g decreases, U will decrease and eventually approach zero. But the answer is B. Could anyone clarify why the answer is B, and not A?


Q2. The earth is approximately 80 times more massive than the moon. The average distance between the earth and the moon is just less than 400,000km. If the radius of the earth is 6370km, the center of gravity of the earth-moon system is located:

A. at the center of the earth
B. just beneath the earth's surface
C. just above the earth's surface
D. exactly between the earth and the moon

The book says this problem can be done by using torque (center of gravity would be the center point of torque). It says since earth is 80 times more massive than the moon, the lever arm for the moon must be 80 times greater. This doesn't make sense to me.

Torque equals force x distance to the center point times sin of theta.
And isn't the force from the earth exactly the same as the force from the moon? F of earth= F of the moon

Using torque equation, it would be m1*a1*r1=m2*a2*r2
Both masses are different (by a factor of 80), but isn't acceleration different for moon and earth since the force would be the same? Then how can r (lever arm for the moon) be 80 times greater?

 

[Ace-Co-A]

For the first question, the U = mgh equation actually doesn't apply. Upon consideration that g decreases as h increases (from what the question tells us), you'll see that this equation fails (i.e., as h goes to infinity, g approaches 1/infinity, which is nonsense (recall from calculus that infinity over infinity does not = 0). What other equations do you recall about gravity?

Recall the relation F = G*[(M*m)/(R^2)] {Eq. 1} where G is the gravitational constant of the universe (G = 6.67 * 10^-11 m^3kg^-1s^-2) and the definition of work (W=F*d) {Eq. 2}. Plug Eq. 1 into Eq. 2 and we get W = PE = G*[(M*m)/(R)] {Eq. 3}. The work required to bring together two objects of masses M and m from some distance R as separated against gravity is given by Eq. 3, and is therefore the potential energy relation needed for the problem. Simple analysis of Eq. 3 shows that the the PE will increase as R decreases.

Hope this helps.

 

[Ace-Co-A]

For question two, I didn't use the torque between the bodies. Instead, I used the definition of the center of mass: McomRcom = m1rcom,1 +m2rcom,2 {Eq. A} In this case, the variables are McomRcom = (Me)(Re) + (m)(rm) {Eq. B} but the distances of the system are judged relative to the center point of one of the two bodies (make it the center of earth for simplicity). Now, Eq. B can be rewritten with the figures given in the text of the problem: (80m+1m)Rcom = (80m)(0km) + (m)(400000km + 6370km) {Eq. C} where m = the mass of the moon.

Solving Eq. C, you get that the Rcom = (406370km)/(81) = 5017km. 5017 km is "just beneath" the surface of the earth (at least relatively, considering that the distance between the two centers is some 406370km). I should state however, that my approach is fairly simple-minded and I may be neglecting a more elegant solution for this problem.

 

[Eastfolding]

You can think of problem 2 like a seesaw.

If you have a 1 kg person on the left side, and an 80 kg person on the right side, where would you put the pivot so that the two balance?

It would need to be 80 times closer to the right side than the left side.

400,000/80 = 5,000, which is less than the radius of the earth.

In reality though, torque has nothing to do with this problem. The answer states it because the equation for balancing a seesaw through torque is fundamentally the same as calculating the center of mass.

 

[Jwinsler7]

For the first question, the U = mgh equation actually doesn't apply. Upon consideration that g decreases as h increases (from what the question tells us), you'll see that this equation fails (i.e., as h goes to infinity, g approaches 1/infinity, which is nonsense (recall from calculus that infinity over infinity does not = 0). What other equations do you recall about gravity?

Recall the relation F = G*[(M*m)/(R^2)] {Eq. 1} where G is the gravitational constant of the universe (G = 6.67 * 10^-11 m^3kg^-1s^-2) and the definition of work (W=F*d) {Eq. 2}. Plug Eq. 1 into Eq. 2 and we get W = PE = G*[(M*m)/(R)] {Eq. 3}. The work required to bring together two objects of masses M and m from some distance R as separated against gravity is given by Eq. 3, and is therefore the potential energy relation needed for the problem. Simple analysis of Eq. 3 shows that the the PE will increase as R decreases.

Hope this helps.

Okay. I think I am having a difficult time grasping the concept. Using the equation 3, where potential energy equals Gm1m2/r, as r (distance between the rocket and Earth is increasing as the rocket moves into deep space) increases, the work required to bring the two objects decreases? And the gravitational potential energy of the rocket increases? I am not sure if I understand that completely..

 

[Eastfolding]

No, potential energy is equal to -GMm/r. At infinity, it is zero, which is the highest potential energy.

 

[Jwinsler7]

No, potential energy is equal to -GMm/r. At infinity, it is zero, which is the highest potential energy.

where did you get the negative sign from?

 

[Eastfolding]

Derive it out of the equation for gravitational force, you'll get a negative sign.

 

[Jwinsler7]

Derive it out of the equation for gravitational force, you'll get a negative sign.

Huh? Derive it out of the equation for gravitational force? What are you talking about?
F= Gm1m2/r^2
PE=W=F*d*cos(theta)=Gm1m2/r

I don't see where you got that negative sign from, unless you used cos180

 

[Ace-Co-A]

Okay. I think I am having a difficult time grasping the concept. Using the equation 3, where potential energy equals Gm1m2/r, as r (distance between the rocket and Earth is increasing as the rocket moves into deep space) increases, the work required to bring the two objects decreases? And the gravitational potential energy of the rocket increases? I am not sure if I understand that completely..

Keep in mind that the gravitational force falls of with the square of the distance. That is, two bodies that are separated by quite a long distance will not "feel" each other very much, and thus the force that pulls them together is fairly low.

Here is where my wording was a little difficult to follow. To do something against gravity, one must pull the body away from the other body. The kinetic energy involved in doing so is the work required to separate them by some distance (let's call it R). The potential energy corresponding to such a tug (that is, the potential energy of the two bodies if they were to move back TOWARDS each other that same distance R in an energetically favorable fashion) must be equal to the work required to separate them by that distance*. So, if we call the distance that we're imagining the rocket is flying AWAY from the Earth R, then the work required to separate them by R km is W=F*R and the PE corresponding to the reverse of that move is PE = [G*M*m)]/R.

*By the conservation of energy: KE - PE = 0

 

[Ace-Co-A]

I'm leaving the negative implicit (as I probably shouldn't) -- potential energy becomes more negative as the distance decreases. (i.e., the absolute value of energy involved increases, as I said before). It is the same principle though. The correct equation is: PE = -[G*(M*m)]/R. (because W = - F*R)

To be explicit, the energetic maximum [where R = infinity and PE = -GMm / (infinity)] is the most unfavorable position. So the potential energy involved in the system decreases (i.e., becomes MORE negative), as the distance separating the bodies decreases.

I've probably thoroughly confused you all now... here's to hoping you can wade through that.

 

[curt656]

I hate to copy and paste from Wikipedia, but it offers a succinct explanation about why gravitational PE is negative in this case...

"However, over large variations in distance, the approximation that g is constant is no longer valid, and we have to use calculus and the general mathematical definition of work to determine gravitational potential energy. For the computation of the potential energy we can integrate the gravitational force, whose magnitude is given by Newton's law of gravitation, with respect to the distance r between the two bodies. Using that definition, the gravitational potential energy of a system of masses m1 and M2 at a distance r using gravitational constant G is:

U= -(Gm1M2)/r

 

[Jengreef]

I'm still a bit confused. If R increases, it feels like PE should get smaller, but not necessarily become more negative. Or are you saying if R increases, some other part of the equation increases more to compensate for increasing R?

 

[Eastfolding]

Huh? Derive it out of the equation for gravitational force? What are you talking about?
F= Gm1m2/r^2
PE=W=F*d*cos(theta)=Gm1m2/r

I don't see where you got that negative sign from, unless you used cos180

You have to integrate the equation, since F changes with R.

So, Work = integral F dx, which is GMm/r^2 dr from zero to R.

This gives you negative GMm/r.

It's really just a convention, you can make believe that maximum gravitational potential energy is infinity instead of zero, and still be conceptually accurate.

 

[Ace-Co-A]

I'm still a bit confused. If R increases, it feels like PE should get smaller, but not necessarily become more negative. Or are you saying if R increases, some other part of the equation increases more to compensate for increasing R?

PE = - [G*(M*m)]/R

Read through what I wrote again. If R increases (the denominator increases, nothing else changes), then the absolute value of the PE decreases. With the negative value included, the potential energy becomes more positive (i.e., approaches zero). If R decreases (the denominator decreases, nothing else changes), then the absolute value of the PE increases. The PE value becomes more negative.

 

[Jwinsler7]

You have to integrate the equation, since F changes with R.

So, Work = integral F dx, which is GMm/r^2 dr from zero to R.

This gives you negative GMm/r.

It's really just a convention, you can make believe that maximum gravitational potential energy is infinity instead of zero, and still be conceptually accurate.

It's just depends on where you set the grav. potential energy to be zero.

Thanks a lot guys for your help, especially 14phillygrad. I get it now.