What Are the Net Forces Acting on the Kuiper Belt Object Ultima Thule?

[OmCheeto]

Homework Statement

[/B]
A Kuiper Belt object, Ultima Thule, rotates once every 15 hours.
It has an estimated density of 1000 kg/m^3.
Ultima, the larger sphere, has a radius of 9000 meters.
Thule, the smaller sphere, has a radius of 6500 meters.
The center of rotation is 4200 meters from the center of Ultima.
G = 6.674E-11 m^3⋅kg−1⋅s−2

a. Find the net forces at the extreme points, A & B. Express the answers as acceleration.
b. Find the ratio of the gravitational to centrifugal forces

x coordinates are in meters


2. Homework Equations

Force of gravity = G M1m2/r^2
Centrifugal force = mrω²
F = ma

The Attempt at a Solution

Step 1
The accelerations can be found by setting F = ma equal to the two forces, and eliminating m.

F = ma = GM1m2/r^2

a = GM1/r^2​

F = ma = mrω²

a = rω²
​

Step 2
I found the volumes and masses of the two spheres
Ultima:

volume = 4/3π(9000^3) = 3.05E+12 m^3
mass = 3.05e12 m^3 * 1000 kg/m^3 = 3.05e15 kg​

Thule:

volume = 4/3π(6500^3) = 1.15E+12 m^3
mass = 1.15e12 m^3 * 1000 kg/m^3 = 1.15e15 kg​

Step 3
Since the object is odd shaped, I broke the gravitational force into two parts:
Net gravitational force = the force from the near sphere + the force from the distant sphere.

Ultima point A

Acc_grav = G(3.05e15 kg)/((9000m)^2) + G(1.15e15 kg)((24500m)^2)
= 0.002516 + 0.000128
= 0.002644 m/s^2​

Thule point B

Acc_grav = G(3.05e15 kg)/((22000m)^2) + G(1.15e15 kg)/((6500m)^2)
= 0.000421 + 0.001817
= 0.002238 m/s^2​

Step 4
Find the accelerations due to rotation from the center of rotation, point C:

distance C to A: 13,200 m
distance C to B: 17,800 m

convert 15 hours into radians/sec
15*60*60 = 54,200 seconds/revolution
2π radians/revolution
ω = (2π rad/rev)/(54200 sec/rev) = 0.000116 radians/second

a = rω²
Ultima point A
Acc_centr = 13200m * ((0.000116/s)^2) = 0.000179 m/s^2

Thule point B
Acc_centrifugal = 17800m * ((000116/s)^2) = 0.000241 m/s^2​

Step 5
Chose a sign convention: Force towards the asteroid is positive.

Step 6
Provide the answers
a.

Ultima point A
Net acc = 0.002644 m/s^2 - 0.000179 m/s^2
net acc = 0.0025 m/s^2

Thule point B
net acc = 0.002238 m/s^2 - 0.000241 m/s^2
net acc = 0.0020 m/s^2​

b.

Ultima
ratio = 0.002644 m/s^2 / 0.000179 m/s^2 = 15

Thule
ratio = 0.002238 m/s^2 / 0.000241 m/s^2 = 9.3​

ps. This is not real homework, so there is no answer in the back of the book. I normally wouldn't post such a trivial problem, but I've gotten indications from 2 different people that my answers may not be correct.

 

[mfb]

Looks good. The approach is fine and all numbers I checked are fine as well.

Quick cross-check: For a sphere with the density of Earth the critical rotation period (centrifugal force = gravity) would be 1.5 hours, the period of a low Earth orbit. 15 hours gives you a safety factor of 100. The density is a factor 5 lower here so our factor goes down to 20. The non-spherical shape removes mass, which makes the factor go down a bit more.

 

[fizzy]

I said your previous answers were not correct but the distances you are now using look good to me.
Seems I'd picked up a false idea about inv. sqr law near to a spherical object. The fact it is exact is quite neat.

Using the masses you calculate, the barycentre distance estimate is 4244km but I'm guessing that NASA have derived their value from observations.

BC_U = ( rT+rU ) * mT / (mT + mU) = 4244km

 

[OmCheeto]

I said your previous answers were not correct

I actually don't remember that.

but the distances you are now using look good to me.

That's strange, as all of the number changed, but the answers look consistent with the changes, so my guess is that my original calculations were also correct.

Seems I'd picked up a false idea about inv. sqr law near to a spherical object. The fact it is exact is quite neat.

Using the masses you calculate, the barycentre distance estimate is 4244km but I'm guessing that NASA have derived their value from observations.

BC_U = ( rT+rU ) * mT / (mT + mU) = 4244km

I've not seen a value posted by NASA. Like you, I calculated the center of mass from the given numbers.
Doh! I see I forgot that comparison.
Distance of center of mass to center of Ultima:

2nd attempt 4700m
hw attempt 4200m
difference: 12%​