Gravity in Tunnel Through Chord of a Planet

[schaefera]

Homework Statement

An object of mass 104 kg moves in a smooth
straight tunnel of length 1540 km dug through
a chord of a planet of mass 3.2 × 10²⁴kg and
radius 1 × 10⁹ m.

Determine the effective force constant
of the harmonic motion.

Answer in units of N/m

Homework Equations

Force of gravity= -(GmM)/(r²)
p=density=mass/volume
Volume of sphere= (4/3)pi*r³

The Attempt at a Solution

Using the equation for force, I know that the gravitational force is only due to the mass of a sphere "inside" of where the point of interest is. Thus, as the mass descends through the tunnel, the force of gravity will decrease. Unlike a problem where the mass goes through the Earth's center, however, this sphere does not decrease to a radius of 0. That is where my issues arise.

Anyway, I get the fact that Mass(inside)=(p)(4/3 pi r³), when I plug this into the force equation, I come out with:

F= -((4/3)Gm*p*pi)*r

The effective constant is that part in front of the r, which represents radius. Is that the value I plug my given values in for (I can find p using the radius and mass of the planet in the problem)? Do I need to account for the fact that the tunnel goes through a chord and not a diameter?

 

[schaefera]

No one?

 

[schaefera]

Bump :)

 

[schaefera]

Must be a tough-y...

 

[schaefera]

hehe

 

[SammyS]

I don't know how I've missed this. I'll look at it & get back to you.

 

[SammyS]

I wouldn't worry too much about the reduced amount of gravitational force in the tunnel relative to at the surface, although that may have a small effect. At the tunnel's center, it is only about 296 meters below the surface. The radius of the planet is 1,000,000,000 meters.

A force constant implies that F = -kx, where x is the distance of the object from the equilibrium position (center of the tunnel), and k is the force constant. (This is analogous to a spring with spring constant k.)

 

[schaefera]

So, k would then be the same whether or not the mass passes through the planet's center?

 

[SammyS]

So, k would then be the same whether or not the mass passes through the planet's center?

Absolutely not. The tunnel is never more than 300 m from the surface at any point. How would k be the same as if the object passed through the planet's center?

 

[schaefera]

Ok well then how do I do it? You see how I tried to resolve force in terms of density in the first post. I can't do that?

 

[SammyS]

Ok well then how do I do it? You see how I tried to resolve force in terms of density in the first post. I can't do that?

That doesn't help. Inside a planet of uniform density, the gravitational force is directly proportional to distance from the planet's center. This tunnel barely penetrates the planet's surface when compared to the planet's radius.

Sketch the situation and define variables. I'll work on a set-up I think may be helpful & post it.

 

[SammyS]

The tunnel is in red, as shown above

The object is located at (x, y₀), where y₀ is the y coordinate of the tunnel.

What is the x component of the gravitational force exerted on the object?

 

[Chronos]

Maximum velocity is the key here. First you need to find acceleration due to gravity of the planet, then calculate velocity achieved by the weight at the midpoint of the chord. Assume it is a ball and rolls friction free through the tunnel.

 

[SammyS]

He only needs to find the force constant. It's pretty basic simply working with components of force.

 

[SammyS]

This problem is quite interesting. It also has some hard to achieve conditions.

The chord (tunnel) is 1540 km in length. The radius of the planet is R=1×10⁶ km. So the chord subtends an angle of 1.5400 milliradians ≈ 0.08824°.

At either entrance to the tunnel, the tunnel makes angle of half the above with respect to the surface of the planet, that's 0.04412°. Imagine a coefficient of friction that's small enough that our object would slide down this slight slope.

As noted earlier, the tunnel intercepts the y-axis about 296.45 meters below the planet's surface. That's about 3.3R/(10 million) from the surface. In other words, if the tunnel intercepts the y-axis at y₀, then y₀ ≈ R[1 - (2.96×10^-7)].

 

[schaefera]

So x-component of the force is going to be Fg(sin(theta))... as in, I find the gravitational force and then multiply by sin(t) (because that is the component across from theta in your diagram). But isn't gravitational force directed in not only the x-direction? Or do I only care about the x-component because the y-component is balanced by the normal of the tunnel on the mass?

I'm still confused how to find the force constant using this method, though.

(Oh, and thanks for all the help! The diagram's great!)

 

[schaefera]

Maximum velocity is the key here. First you need to find acceleration due to gravity of the planet, then calculate velocity achieved by the weight at the midpoint of the chord. Assume it is a ball and rolls friction free through the tunnel.

Couldn't I just used energy considerations, knowing that U(i)=U(f)+K(f)? That seems easier to me, if the question had been finding maximum velocity.

 

[SammyS]

I suppose you can do it that way, but the force way is really simple.

 

[SammyS]

So x-component of the force is going to be Fg(sin(theta))... as in, I find the gravitational force and then multiply by sin(t) (because that is the component across from theta in your diagram). But isn't gravitational force directed in not only the x-direction? Or do I only care about the x-component because the y-component is balanced by the normal of the tunnel on the mass?

I'm still confused how to find the force constant using this method, though.

(Oh, and thanks for all the help! The diagram's great!)

Yes, F_g has a large component in the y direction, but the object can't move in that direction. A normal force will cancel that component.

The whole tunnel is so near the surface that F_g ≈ GmM/R² (in a direction toward the center of the planet), where R is the radius of the planet.

So as you said the x component is F_x = -F_gsin(θ).

Notice that sin(θ) = x/R

Put these together & compare to F=-kx.

BTW: Just out of curiosity, What level course is this for?

 

[schaefera]

It's for an AP Physics C course- the mechanics portion. Not college, per se, but I certainly hope college isn't too much more difficult for the intro level!

 

[SammyS]

$$F_G\cdot\sin(\theta)=-G\frac{m\,M}{R^2}\cdot\frac{x}{R}$$