An asteroid flying past the moon

In summary: I need to take the difference in the potential energy in order to find the kinetic energy?Yes that is one correction. And I hope that you have taken the difference in the potential energy (near the moon and near the ground) instead of only the potential energy at the position.The conservation of energy equation is K1 + U1=K2 + U2. Why did I have to take the difference in the potential energy though?Well, if you want to find K2, you need to add K1 and U1-U2. Hence you would require the difference in the potential energy.That means that the mechanical energy equals to the kinetic energy
  • #1
*Jessica*
11
0
I am not that great in physics, but I put a lot of effort in trying to learn it. I would really appreciate any help on this problem, since I am not even sure where to start.

An asteroid is about to fly past the moon (about 384,000km away) and is on a collision course with the earth. The mass of the asteroid is about 1.45e20 kg. It approaches Earth on a direct impact course with a velocity of about 2.0m/s. You are part of an elite team of experts assigned to find a way to deflect the path of the asteroid. Note: Mass of the Earth is about 6e24 kg, and the radius is 6400km.
1) Ignoring the effects of air-resistance, if the total kinetic energy of the asteroid was released in an explosion when it impacts the surface of the earth, what is the TNT equivalent of that energy? 1 ton TNT = 4.184e9 J. 1 megaton TNT = 1Mt = 1e6 t of TNT = 4.184e15 J.
2) If all of this energy can be used to power 7W LED light bulbs for a year, how many such bulbs can you power in total?
3) Your team devised a plan to insert and detonate a nuclear device in the center of the asteroid. Suppose this was done in a timely fashion, so that the asteroid has just past the moon. Your team had calculated that about 60% of the energy will go into splitting the asteroid into two halves of the same mass and that the two halves will begin to move apart along a direction perpendicular to the initial velocity vector with the hope that the two parts will miss the earth. What is the total energy needed in the detonation? How many TNT equivalents do we need? If a typical nuclear bomb yields 50Mt of energy, how many will you need? (For this part, ignore the effects of gravity.)
I guess I need to use kinematics and gravitation equations.Any help will be appreciated.
 
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  • #2
Hello *Jessica*

You could start with calculating the mechanical energy possessed by the asteroid. What would be the total energy it possesses given its position relative to Earth and its speed?

By the way I need to clarify this; is the speed 2.0 m/s when it just passes the moon or when it is just about to strike the Earth?
 
  • #3
I find energy using this equation, right?

E=1/2 mv^2 + (-(Gm_1m_2)/r^2)

I am not sure about the velocity. This is the exact wording.
 
  • #4
*Jessica* said:
I find energy using this equation, right?

E=1/2 mv^2 + (-(Gm_1m_2)/r^2)

I am not sure about the velocity. This is the exact wording.

Check your potential energy expression.

I guess you'll need to assume that speed is near the moon and see whether we get the prescribed answer. If you don't then try the other assumption.
 
  • #5
Oh yes! The radius is not squared in the potential energy equation. So I get E= -9.067e27 J. How do I find now the kinetic energy it released?
 
  • #6
*Jessica* said:
Oh yes! The radius is not squared in the potential energy equation. So I get E= -9.067e27 J. How do I find now the kinetic energy it released?

Yes that is one correction. And I hope that you have taken the difference in the potential energy (near the moon and near the ground) instead of only the potential energy at the position.

Well, once you get the total energy, use law of conservation of it. What would be the total kinetic energy near the ground?
 
  • #7
Uhm no, I didn't take the difference. So my potential energy near the Earth will have the radius of the Earth in the denominator and the potential energy near the moon will have 384e6 m (or this number plus the radius of the Earth)?
The conservation of energy equation is K1 + U1=K2 + U2. Why did I have to take the difference in the potential energy though?
 
  • #8
*Jessica* said:
Uhm no, I didn't take the difference. So my potential energy near the Earth will have the radius of the Earth in the denominator and the potential energy near the moon will have 384e6 m (or this number plus the radius of the Earth)?
The conservation of energy equation is K1 + U1=K2 + U2. Why did I have to take the difference in the potential energy though?

Well, if you want to find K2, you need to add K1 and U1-U2. Hence you would require the difference in the potential energy.
 
  • #9
That means that the mechanical energy equals to the kinetic energy, since the equations are the same. Are the equations for potential energy correct?
 
  • #10
Or do these potential energies cancel out each other and all we are left with is 1/2(mv^2)?
 
  • #11
Find U1-U2 and see for yourself.

So then,what would be the final kinetic energy?
 
  • #12
1/2(mv^2). Thank you!
Can you help with part 3, please?
 
  • #13
*Jessica* said:
1/2(mv^2). Thank you!
Can you help with part 3, please?

Hey could you show your working to obtain the final KE to just be sure?

As for the third part, Since they have told us to ignore gravitational effects of the earth, Just imagine what the trajectories of the two pieces look like. Have you got the picture?
 
  • #14
Since we need to find the kinetic energy when the asteroid hits the Earth, the potential energy is 0, leaving us with only the equation for kinetic energy.

The two pieces will just move in the opposite directions. Do we use the conservation of momentum here?
 
  • #15
*Jessica* said:
Since we need to find the kinetic energy when the asteroid hits the Earth, the potential energy is 0, leaving us with only the equation for kinetic energy.

The two pieces will just move in the opposite directions. Do we use the conservation of momentum here?

No, the potential energy at the surface is not zero. Just calculate U1-U2. It is not zero. (think about it, according to you argument, an asteroid falling to Earth won't speed up)

Draw the trajectory of the pieces to get an idea.
 
  • #16
K2=0.5(1.45e20)(2.0)^2 + (-(G(1.45e20)(6e24))/6.4e6)-(G(1.45e20)(6e24))/3.84e8)
 
  • #17
*Jessica* said:
K2=0.5(1.45e20)(2.0)^2 + (-(G(1.45e20)(6e24))/6.4e6)-(G(1.45e20)(6e24))/3.84e8)

Shouldn't the change in potential energy be Umoon-Uground?
 
  • #18
Yes, you are right. I was just super tired, now I see it.
And for part 2 I simply multiply 7W by the number of seconds in a year and then divide the number of joules by that number.
 
  • #19
The answer to part 2 seems correct
Hope you understood what to do with part 3
 
  • #20
Not really.. I don't understand the 60% part. It's 60% of what energy? The kinetic one I calculate in part 1?
 
  • #21
No, I think they mean 60% of the nuclear blast's energy.
 
  • #22
Sunil Simha said:
No, I think they mean 60% of the nuclear blast's energy.

Yes...

And Jessica,

Once you calculate it, you can try conserving energy (mechanical) of the asteroid system. Conserving momentum will be brain bashing as it is not conserved for asteroid system, but for asteroid + Earth system. (Why ?)
 

What is an asteroid?

An asteroid is a small rocky object that orbits around the sun. They are leftovers from the formation of the solar system and can range in size from a few feet to hundreds of miles in diameter.

How often do asteroids fly past the moon?

Asteroids fly past the moon quite frequently, with an estimated 50-100 passing within the moon's distance every year. However, only a small percentage of these are considered close approaches, meaning they come within 4 million miles of the moon.

Is there a risk of an asteroid hitting the moon?

Yes, there is a small risk of an asteroid hitting the moon. However, the moon's gravity and lack of atmosphere make it a less likely target for asteroid impacts compared to Earth. The moon also has a much smaller surface area, making the chances of a direct hit even lower.

What would happen if an asteroid hit the moon?

If a large enough asteroid were to hit the moon, it could potentially create a crater and alter the moon's surface. However, the moon's gravity is much weaker than Earth's, so the effects of an impact would be less severe.

Can we predict when an asteroid will fly past the moon?

Yes, scientists are able to predict when an asteroid will fly past the moon using advanced telescopes and computer simulations. This allows them to track potentially hazardous asteroids and determine if they pose a risk to Earth or the moon.

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