Gravitational Potential In the Field Caused by Two Masses

[BrandonInFlorida]

Homework Statement

An 800-kg mass and a 600-kg mass are separated by 0.25 meter. (a) What is the gravitational field strength due to these masses at a point 0.20 meter from the 800-kg mass and 0.15 meter from the 600-kg mass? (b) What is the gravitation potential at this point due to these same masses?


Answers:

a. 2.2 x 10 to the -6 m/s2

b. -5.3 x 10 to the -7 joules/kg

Relevant Equations

U = GM1M2/R

I got (a) but have no idea about b. Potential fields aren't just additive all the time are they?

 

[haruspex]

Potential fields aren't just additive all the time are they?

They are. Have you tried just adding them?

 

[kuruman]

I got (a) but have no idea about b.

Can you explain how you got (a)?

I don't understand where the point of interest lies. I assume that the given 0.25 m separation is center-to-center. If so, the point of interest doesn't fit on the line joining the centers because it is 0.20 m from the 800-kg and 0.15 from the 600 kg sphere, and 0.25 < 0.20 + 0.15. If the point of interest is not on the center-to-center line, we need more information about the point's location to do part (a).

 

[haruspex]

we need more information about the point's location to do part (a).

Why? The triangle is completely determined by the three distances.

 

[Delta2]

Yes well I think @kuruman meant to say that we need to do additional work, that is calculate the angles of the triangle, and do vectorial add of the two contributions.

 

[haruspex]

Yes well I think @kuruman meant to say that we need to do additional work, that is calculate the angles of the triangle, and do vectorial add of the two contributions.

Since @BrandonInFlorida claims to have got that part right, I presume he did all that.

 

[Delta2]

Since @BrandonInFlorida claims to have got that part right, I presume he did all that.

yes well and it seems to me he did the hard part but he couldn't do the easy part that needs just scalar addition. But ok I guess he didn't know that potentials add just like two scalars.

 

[PeroK]

I got (a) but have no idea about b. Potential fields aren't just additive all the time are they?

If forces are additive and the gradient of the potential then we have:
$$\vec F = \vec F_1 + \vec F_2 = -\nabla U_1 - \nabla U_2 = -\nabla(U_1 + U_2)$$
Hence, $U = U_1 + U_2$ will do nicely as the potential for the system.

 

[kuruman]

Yes well I think @kuruman meant to say that we need to do additional work, that is calculate the angles of the triangle, and do vectorial add of the two contributions.

Actually, it was a final version of an initial posting that edited late at night. Here is why. Initially, the numbers didn't sit well with me and they still don't. The spheres are not point masses otherwise the problem would not have said "spheres". I believe that to answer the questions, the actual density of the spheres would be needed. Even if the spheres were made of osmium, the densest element with ρ = 22.56×10³ kg/m³, the 800-kg sphere would have a radius of 20.4 cm and the 600-kg sphere a radius of 18.5 cm. Not only the point of interest is inside both spheres no matter what their composition, but also the spheres would overlap. That would make the calculation interesting but not undoable.

Eventually, I decided to omit all that as too much of an aside and go with the point sphere model (I even calculated that the spheres could be made of neutron star material with a radius of 11 microns). it seems that I butchered the original version when I changed its scope.

 

[Steve4Physics]

Relevant Equations:: U = GM1M2/R

Hi @BrandonInFlorida . I’d like to highlight some points in addition to what the others have said.

Your formula “U = GM1M2/R should be “U = -GM1M2/R”. Gravitational potential energy is negative in this context.

(For information, the formula applies to two spherically symmetric (or point), non-overlapping masses, where R is their centre-to-centre distance.)

But, even corrected, this is the wrong formula for this question!

In your formula, U is the gravitational potential energy of a system of 2 masses. But you are not being asked to find U. You are being asked for the gravitational potential (say ‘V’) at a particular point.

The gravitational potential at a point P outside of a single spherically symmetric (or point) mass, M, is
$V = - \frac {GM}{r}$ where r is the distance from M's centre to P.

At a given point you can simply add potentials arising from different masses to get the total potential at that point.

 

[BrandonInFlorida]

Can you explain how you got (a)?

I don't understand where the point of interest lies. I assume that the given 0.25 m separation is center-to-center. If so, the point of interest doesn't fit on the line joining the centers because it is 0.20 m from the 800-kg and 0.15 from the 600 kg sphere, and 0.25 < 0.20 + 0.15. If the point of interest is not on the center-to-center line, we need more information about the point's location to do part (a).

I figured out that it was a right triangle. That was the additional information I needed, and I did get the answer at the back of the book.

 

[BrandonInFlorida]

Hi @BrandonInFlorida . I’d like to highlight some points in addition to what the others have said.

Your formula “U = GM1M2/R should be “U = -GM1M2/R”. Gravitational potential energy is negative in this context.

(For information, the formula applies to two spherically symmetric (or point), non-overlapping masses, where R is their centre-to-centre distance.)

But, even corrected, this is the wrong formula for this question!

In your formula, U is the gravitational potential energy of a system of 2 masses. But you are not being asked to find U. You are being asked for the gravitational potential (say ‘V’) at a particular point.

The gravitational potential at a point P outside of a single spherically symmetric (or point) mass, M, is
$V = - \frac {GM}{r}$ where r is the distance from M's centre to P.

At a given point you can simply add potentials arising from different masses to get the total potential at that point.

Yes, I stated this sloppily. Everything you've said is true. My issue was that even after I wrote this post when I found out that people say the potentials from the individual masses are additive, it didn't seem obvious to me. I thought you could get a case where the potentials oppose each other rather than just adding. Finally, I decided that if you look at it as the two individual masses each doing work on an object being brought in from infinity, I could see that it would be additive.

 

[BrandonInFlorida]

If forces are additive and the gradient of the potential then we have:
$$\vec F = \vec F_1 + \vec F_2 = -\nabla U_1 - \nabla U_2 = -\nabla(U_1 + U_2)$$
Hence, $U = U_1 + U_2$ will do nicely as the potential for the system.

Good point!