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Asteroid: Potential/Kinetic Energy, Angular Momentum

by Emanresu12
Tags: angular, asteroid, energy, momentum
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Emanresu12
#1
Nov18-09, 11:57 PM
P: 1
1. The problem statement, all variables and given/known data

Suppose an asteroid of mass 5e20 kg is nearly at rest outside the solar system, far beyond Pluto. It falls toward the Sun and crashes into the Earth at the equator, coming in at an angle of 30 degrees to the vertical as shown, against the direction of rotation of the Earth. It is so large that its motion is barely affected by the atmosphere.

(a) Calculate the impact speed.
(b) Calculate in hours the change in the length of a day due to the impact.

Note: Assume that the mass of the asteroid does not significantly change the moment of inertia of the Earth. This is a valid assumption except for the very largest of asteroids.

2. Relevant equations

Gravitational Constant G: 6.67e-11 m^3/kg/s^2

Mass of Earth (M_e): 6e24 kg
Mass of Sun (M_s): 2e30 kg
Mass of Asteroid (m_a): 5e20 kg

Distance from Earth to Sun (r1): 1.5e11 m
Distance from Sun to Edge of Solar System(r2): 200 AU (3e13 m) ~not sure if this is a good approximation or not.
Radius of Earth(r3): 6.4e6

Angle of Asteroid Impact from Earth's Vertical: 30 degrees

Period of Earth's Rotational Angular Momentum (T): 86400 seconds

Velocity of Earth (v_e): 464 m/s

Kinetic Energy: K=(1/2)mv^2
Potential Gravitational Energy: U=GMm/r
Rotational Angular Momentum: Lrot=Iw
Moment of Inertia of a Sphere: (2/5)MR^2
Moment of Inertia: mr^2
Angular Velocity: 2pi/T
Translational Angular Momentum: Ltrans=mvr
Centripetal Force: Fc=mv^2/r

3. The attempt at a solution

Part A
I apply Conservation of Energy and assume that the kinetic energy that the Asteroid gains from its initial state at rest at the edge of the solar system is equal to the gravitational potential energy between the Asteroid and the Sun:

0.5(m_a)v^2)G(M_s)(m_a)/(r2-r1)

I solved for v to get the impact speed:

v=((2G)(M_s)/(r2-r1))^(1/2)

Which yields a velocity of 2990 m/s. However, this is incorrect. The reasons I can think that this may be wrong are:

(1) Bad estimate of the edge of the solar system ~this seems fairly subjective and I haven't been able to find any definite, "hard" numbers for this.
(2) I need to account for the gravitational force between the Earth and Asteroid. Initially, compared to the mass of the Sun and the large distance involved, I would assume that this is of a negligible force.

Part B
Even though I believe I still need the impact velocity to calculate this, I've already gone ahead and thought of a possible solution. I counted the Asteroid as an external force and considered the Earth already rotating with an initial angular velocity of:

w=2pi/T

Yielding w=7.27e-5 seconds. I also considered the Earth's moment of inertia to be:

Irot=(2/5)(Me)(r3)^2

Yielding Irot=7.68e31 kgm^2. So, Earth's rotational angular momentum would be:

Lrot=Irotw

Yielding Lrot=5.58e27 Nms. Next I solve forEarth's translational angular momentum so as to find the total momentum of the system:

Ltrans=(m_e)(v_e)(r3)

Yielding Ltrans=1.78e34 Nms. The total angular momentum would be roughly equal to Ltrans since it's a much larger magnitude than Lrot. By the conservation of momentum, this total would remain constant both before and after the asteroid hits. However from this point I have no idea how to factor in how the asteroid would change the total angular momentum, which would change the rotational angular momentum with the translational angular momentum remaining constant, resulting in the change of period T of Earth's day.

Thanks for any help!
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
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kuruman
#2
Nov19-09, 03:29 PM
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kuruman's Avatar
P: 3,443
The potential energy change is

[tex]-\frac{GMm}{r_f}- \large( -\frac{GMm}{r_i}\large)=GMm(\frac{1}{r_i}-\frac{1}{r_f})[/tex]

It looks like you have

[tex]\frac{GMm}{r_i-r_f}[/tex]

in your expression.


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