Acceleration due to gravity of rod

[oddjobmj]

Homework Statement

Consider a thin rod of mass M and length L. Determine the gravitational potential at a point which is a distance r from the center of the rod and which lies on the midplane of the rod.

Now suppose M is equal to the mass of the Earth and L is equal to the radius of the Earth.

Determine the acceleration due to gravity (of the rod) at a distance r = 0.86 L from the center of the rod.

Homework Equations

$\Phi$(r)=$\int$^L/2_-L/2$\frac{-G(M/L)}{sqrt(z^2+r^2)}$dz

U=mgh

The Attempt at a Solution

I calculate the gravitational potential to be -6.91657*10^7 Joules.

mgh=-6.91657*10^⁷ Joules

g=$\frac{-6.91657*10^7}{m*h}$

The distance from the rod is r=.86*L but what is m? If I ignore m I get -12.62 m/s^2 which is incorrect.

Suggestions? Thank you!

 

[ehild]

Homework Statement

Consider a thin rod of mass M and length L. Determine the gravitational potential at a point which is a distance r from the center of the rod and which lies on the midplane of the rod.

Now suppose M is equal to the mass of the Earth and L is equal to the radius of the Earth.

Determine the acceleration due to gravity (of the rod) at a distance r = 0.86 L from the center of the rod.

Homework Equations

$\Phi$(r)=$\int$^L/2_-L/2$\frac{-G(M/L)}{sqrt(z^2+r^2)}$dz

Correct so far... What did you get for the integral?

U=mgh

mgh is the gravitational potential energy of a mass m at height h above the Earth surface. It is irrelevant here.

The [STRIKE]potential[/STRIKE] force in the gravitational field of the rod is the force acting on unit mass. It depends on the mass of the rod and contains the gravitational constant G.

I calculate the gravitational potential to be -6.91657*10^7 Joules.

mgh=-6.91657*10^⁷ Joules

g=$\frac{-6.91657*10^7}{m*h}$

The distance from the rod is r=.86*L but what is m? If I ignore m I get -12.62 m/s^2 which is incorrect.

Suggestions? Thank you!

That is all wrong. Determine the integral, and substitute r=0.86 L, M= M(Earth) and L= R(Earth)

Edit:
The [STRIKE]potential [/STRIKE] negative gradient of the potential is the magnitude of the gravitational force on unit mass, and the gravitational force exerted on mass m is F=-gradΦ m, and F=ma . The magnitude of the gravitational acceleration is equal to the gradient of the potential at the given point.

ehild

 

[haruspex]

The potential in the gravitational field of the rod is the force acting on unit mass.

No, that's the field, the grad of the potential.

I calculate the gravitational potential to be -6.91657*10^7 Joules.

Gravitational potential has dimension L²T^-2. So J/kg here.

mgh=-6.91657*10^7 Joules

No, you cannot equate the potential to mgh. First, you don't want the m at all. The potential is a property of the field, independently of any mass placed in it. Ditching the m at least makes the dimensions right. But neither is it equal to gh. That would be the potential difference when moving with or against a uniform field for a distance h. The potential is zero at infinity and negative everywhere else.
For the acceleration you need to differentiate the potential wrt r.

 

[ehild]

No, that's the field, the grad of the potential.

Huh, I made a big mistake. Thank you, haruspex that you warned me.

ehild

 

[oddjobmj]

Ah, not sure what I was thinking. It was a long night. Thank you!

Evaluating the integral I get:

$\frac{-2GMsinh^{-1}(L/2r)}{L}$

Taking the derivative of that with respect to r I get:

$\frac{2GMr}{(r^2)^{3/2}sqrt{L^2/r^2+4}}$

Plugging in my known values I get 11.45 m/s²

Thank you both for all your help!