• Support PF! Buy your school textbooks, materials and every day products Here!

Asteroid collision

  • Thread starter karobins
  • Start date
  • #1
7
0

Homework Statement


Suppose an asteroid of mass 1 1019 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 (Figure 9.71). It is so large that its motion is barely affected by the atmosphere.




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



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.


Homework Equations



mass earth: 6e24
mass of asteriod: 1e19
r1 = distance from sun to beyond pluto ~ 5.89e12
r2 = distance from sun to earth = 1.5e11

V = GMm / R
ω = 2π /T
I = mr^2

The Attempt at a Solution



I've sat here and stared at it for about an hour now and I have absolutely no idea.
 

Answers and Replies

  • #2
LowlyPion
Homework Helper
3,090
4

Homework Statement


Suppose an asteroid of mass 1 1019 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 (Figure 9.71). It is so large that its motion is barely affected by the atmosphere.

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

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.

Homework Equations



mass earth: 6e24
mass of asteriod: 1e19
r1 = distance from sun to beyond pluto ~ 5.89e12
r2 = distance from sun to earth = 1.5e11

V = GMm / R
ω = 2π /T
I = mr^2

The Attempt at a Solution



I've sat here and stared at it for about an hour now and I have absolutely no idea.
First of all I would want to figure the Kinetic energy that the asteroid had gained in falling toward the sun. 1/2 mV2 which you can figure by figuring its change in potential energy wrt the sun right?

That should set up the next part of figuring the change in angular momentum.

This you might find by figuring the angular momentum of the earth and then subtracting the angular momentum of the asteroid relative to the center of the earth and figuring how that changes the final ω.
 
  • #3
174
0
Sorry for interrupting ,
Could You please attach the figure(sketch)?
thanks.
 
  • #4
7
0
here is the figure
 

Attachments

  • #5
Yeah I'm stuck on this problem too...
 
  • #6
Ok, do NOT ask me why, but I believe that part A is 4.4e4 m/s. Hope that is the answer! :)
 
  • #7
21
0
gooooooood job
that's the ans.......
but don't know why
haha
 
  • #8
LowlyPion
Homework Helper
3,090
4
Ok, do NOT ask me why, but I believe that part A is 4.4e4 m/s. Hope that is the answer! :)
Seems close, perhaps you used different numbers?

Figuring that the asteroid is at rest with 0 potential when it starts, then you have

0 = 1/2 m*v2 + (- m*μs /r)

where μs is the heliocentric gravitational constant of 1.32 x 1020 m3/s2

I used r from the sun of 1.5*1011 m
 
  • #9
3
0
In order to figure out the velocity at impact, you use the energy principle (K + U)i = (K + U)f. Initially, the asteroid's velocity is zero, and it is far enough away for us to consider the gravitational potential energy zero as well. In the final state, the asteroid is nearly touching the earth's surface, so it has gravitational potential energy relative to the sun, AND the earth. So, 0i = (.5mv^2- G*M_earth*M_asteroid/r_1 -G*M_sun*M_asteroid/r_2)f, where r_1 = radius of earth, and r_2 = distance between the earth and sun. Knowing everything except v, you can solve for it.

G = 6.7e-11
M_asteroid = 1e19
M_earth = 6e24
M_sun = 2e30
r_1 = 6.4e6
r_2 = 1.496e11

I got 43784.34 m/s for the impact velocity.

In part b, the velocity that changes the rotational angular momentum is sin(theta)*(impact velocity), but since they ask for the change in period IN HOURS, you need to convert sin(theta)*v_impact into m/hr, then you use the conservation of angular momentum. Since asteroid impacts in the opposite direction of the angular momentum of the earth, you have -sin(theta)*v_impact(in m/hr)*r_1 + 2/5*M_earth*r_1^2*2*pi/24 initially, and that is equal to the final state: 2/5*M_earth*r_1^2*2*pi/T, where T is your new period. Solve for T, and you should get 24.00470466 hrs or something close, subtract 24 from that, and you have the change in length of one day. Boiler UP!!!
 
  • #10
3
0
Note: different masses of the asteroid will result in different changes in the length of the day
 
  • #11
3
0
Correction!!!! you should multiply by the mass of the asteroid in part b where it says -sin(theta)*v_impact*r_1. So it should be -sin(theta)*v_impact*r_1*m_asteroid. Sorry if it caused problems
 

Related Threads for: Asteroid collision

  • Last Post
Replies
2
Views
598
  • Last Post
Replies
0
Views
230
Replies
2
Views
1K
  • Last Post
Replies
12
Views
5K
  • Last Post
Replies
5
Views
12K
Replies
6
Views
10K
Replies
3
Views
5K
Replies
16
Views
768
Top