Derivative of Newton's Law of Gravitation

In summary, the Newton's Law of Gravitation states that the magnitude of force (F) exerted by a body of mass (m) on a body of mass (M) is equal to (GmM)/(r^2), where G is the gravitational constant and r is the distance between the bodies. When finding the derivative dF/dr, m and M are treated as constants, as the derivative is taken with respect to r and all other variables are held constant. The resulting derivative is -2GmM/r^3, indicating that as the distance between the bodies decreases, the force of attraction between them increases. The negative sign signifies that the force is attractive.
  • #1
Bear_B
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Homework Statement


Newton's Law of Gravitation says that the magnitude (F) of the force exerted by a body of mass (m) on a body of mass (M) is

F = (GmM)/(r^2)

where G is the gravitational constant and r is the distance between the bodies.

Find dF/dr and explain it's meaning. What does the minus sign indicate?

Homework Equations


none

The Attempt at a Solution



using implicit differentiation:
dF/dr = d/dr [(GmM)/r^2]
= [(r^2)d/dr(GmM) - (GmM)d/dr(r^2)] / (r^4)
= [(r^2)G(d/dr(mM)) - (2rGmM)] / (r^4)
= [(r^2)G(m * dM/dr + M * dm/dr) - 2rGmM] / (r^4)

Ok, I know this can be simplified, but that seemed even more convoluted when I did that. I'm really not sure what to do from here. I expected the derivative to be something a little more intuitive and to result in a negative expression based on the question. I would expect dF/dr to equal a negative expression and the significance of the minus sign to be indicative of a force of attraction since the derivative is with respect to distance and the distance is decreasing. Please help me with the derivative, because If I can get that right, then I just need to express what's going on in words.
 
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  • #2
Hi Bear_B! :wink:

(try using the X2 tag just above the Reply box :wink:)
Bear_B said:
Newton's Law of Gravitation says that the magnitude (F) of the force exerted by a body of mass (m) on a body of mass (M) is

F = (GmM)/(r^2)

where G is the gravitational constant and r is the distance between the bodies.

Find dF/dr and explain it's meaning. What does the minus sign indicate?

dF/dr = d/dr [(GmM)/r^2]
= [(r^2)d/dr(GmM) - (GmM)d/dr(r^2)] / (r^4)

ooh, so complicated! :cry: no, m and M are constants.

Try again. :smile:
 
  • #3
Ok, I had seen that result worked out elsewhere when I couldn't get on here. Can you please tell me why m and M are constants? I thought the equation to determine the gravitational force was for an unknown mass m and an unknown mass M at a unknown distance r from each other...knowing that m and M are constants makes the differentiation simple, I just want to understand why I am treating m and M as constants. Thanks
 
  • #4
F = GmM/r2
dF/dr = GmM [ d/dr (r-2)]
= GmM (-2r-3)
= GmM (-2/r3)
dF/dr = -2GmM / r3

ok, so here is the answer, but why are m and M constants?
 
  • #5
Because they have nothing to do with r. They are masses.
 
  • #6
Bear_B said:
Can you please tell me why m and M are constants? I thought the equation to determine the gravitational force was for an unknown mass m and an unknown mass M at a unknown distance r from each other...knowing that m and M are constants makes the differentiation simple, I just want to understand why I am treating m and M as constants. Thanks

Hi Bear_B! :smile:

(just got up :zzz: …)

Yes, m and M are variables just like r,

but they are independent variables (they can be varied separately, without affecting the others).

So when the question asks "Find dF/dr", it means differentiate with respect to r, keeping all other variables constant. :smile:
 
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  • #7
tiny-tim said:
"Find dF/dr", it means differentiate with respect to r, keeping all other variables constant. :smile:

Thanks. Now I understand the concept that the Force (F) is being taken with respect to distance (r) determining the rate of change of F as r changes for two objects with mass m and M. Since dF/dr is asking how F changes when r changes it has to assume that the m and M are constants unless otherwise indicated. I think I get it now. I just wanted a conceptual understanding of the problem beyond how to do the problem mechanically (because that is the easy part).

^^ This is my (basic) understanding of why the derivative assumes m and M as constants when differentiating with respect to r.
 

1. What is Newton's Law of Gravitation?

Newton's Law of Gravitation is a fundamental law of physics that describes the force of gravitational attraction between two objects. It states that every object in the universe attracts every other object with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

2. What is the derivative of Newton's Law of Gravitation?

The derivative of Newton's Law of Gravitation is a mathematical expression that represents the rate of change of the gravitational force with respect to either the mass or the distance between two objects. It is derived from the original law by taking the limit as the distance between the objects approaches zero.

3. Why is the derivative of Newton's Law of Gravitation important?

The derivative of Newton's Law of Gravitation is important because it allows us to understand how the gravitational force changes as the mass or distance between two objects changes. This is crucial for understanding the dynamics of celestial bodies and predicting their movements.

4. How is the derivative of Newton's Law of Gravitation calculated?

The derivative of Newton's Law of Gravitation is calculated using calculus, specifically the derivative rules for products and inverses. By taking the derivative of the original law, we can determine the mathematical expression for the derivative of the gravitational force.

5. What are some real-world applications of the derivative of Newton's Law of Gravitation?

The derivative of Newton's Law of Gravitation has many real-world applications, such as predicting the orbits of planets and satellites, understanding the tides on Earth, and calculating the trajectories of spacecraft. It is also used in gravitational wave astronomy to detect and study cosmic events such as black hole mergers.

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