OmCheeto
Gold Member
 2,023
 2,366
1. The problem statement, all variables and given/known data
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 = 6.674E11 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. Relevant equations
Force of gravity = G M1m2/r^2
Centrifugal force = mrω²
F = ma
3. The attempt at a solution
Step 1
The accelerations can be found by setting F = ma equal to the two forces, and eliminating m.
I found the volumes and masses of the two spheres
Ultima:
Thule:
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
Thule point B
Step 4
Find the accelerations due to rotation from the center of rotation, point C:
Step 5
Chose a sign convention: Force towards the asteroid is positive.
Step 6
Provide the answers
a.
b.
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.
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 = 6.674E11 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. Relevant equations
Force of gravity = G M1m2/r^2
Centrifugal force = mrω²
F = ma
3. 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
Step 2a = GM1/r^2
F = ma = mrω²a = rω²
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
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
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
= 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
= 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
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
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
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.
Attachments

32 KB Views: 167