How will the equation for gravitational force be changed?

[Edel Crine]

Homework Statement

There is a gravitational force between a particle and a uniform rod.
m is a mass of particle, M is a mass and L is a length of the rod. d is the distance between the head of the rod and particle.
If the d>>L, how does the equation for the gravitational force of the rod to a particle will be changed?

Relevant Equations

F = GmM/d(d+L)

At first, I thought when the rod goes really far away from the particle, then L would approach to zero in a particle's reference view. As a result, the equation will be GmM/d^2. However, I just thought that L just remain as it is regardless of change in d, but not sure...

 

[Delta2]

You are correct but for the wrong reason. What you are saying might have some intuitive sense but it is mathematically incorrect or inaccurate. What really happens is that when d becomes very large in comparison to L then the ratio $\frac{L}{d}$ tends to zero.

How can you take advantage that for $d>>L$ it will be $\frac{L}{d}\approx 0$ to prove mathematically that the force will tend to $F\approx G\frac{mM}{d^2}$?

 

[Edel Crine]

You are correct but for the wrong reason. What you are saying might have some intuitive sense but it is mathematically incorrect or inaccurate. What really happens is that when d becomes very large in comparison to L then the ratio $\frac{L}{d}$ tends to zero.

How can you take advantage that for $d>>L$ it will be $\frac{L}{d}\approx 0$ to prove mathematically that the force will tend to $F\approx G\frac{mM}{d^2}$?

Ummmm... using limit...?
lim d->∞ (L/d) = L/∞ = 0...?

 

[Delta2]

You are just using limits to prove that $$lim_{d \to +\infty}\frac{L}{d}=0$$ which is fine but I didn't ask you to do this. Essentially I asked you to prove that $$\lim_{\frac{L}{d} \to 0}G\frac{mM}{d(d+L)}=G\frac{mM}{d^2}$$

 

[Edel Crine]

You are just using limits to prove that $$lim_{d \to +\infty}\frac{L}{d}=0$$ which is fine but I didn't ask you to do this. Essentially I asked you to prove that $$\lim_{\frac{L}{d} \to 0}G\frac{mM}{d(d+L)}=G\frac{mM}{d^2}$$

I'm so sorry... little stuck...

 

[Edel Crine]

You are just using limits to prove that $$lim_{d \to +\infty}\frac{L}{d}=0$$ which is fine but I didn't ask you to do this. Essentially I asked you to prove that $$\lim_{\frac{L}{d} \to 0}G\frac{mM}{d(d+L)}=G\frac{mM}{d^2}$$

Oh, if I divide both numerator and denominator by d^2, then
(GmM/d^2)*(1/(1+L/d))
=GmM/d^2)*(1/(1+0))
=(GmM/d^2)*1
=GmM/d^2
?

 

[Delta2]

Well now you have essentially proven it correctly :D.

You just have to learn to use $LATEX$ though :D

 

[Edel Crine]

Well now you have essentially proven it correctly :D.

You just have to learn to use $LATEX$ though :D

I really appreciate! That makes a lot more sense to me! Yes, learning LATEX would be much easier to show my works... Anyway, you have a great day!