An asteroid flying past the moon

[*Jessica*]

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.

 

[Sunil Simha]

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?

 

[*Jessica*]

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.

 

[Sunil Simha]

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.

 

[*Jessica*]

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?

 

[Sunil Simha]

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?

 

[*Jessica*]

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?

 

[Sunil Simha]

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.

 

[*Jessica*]

That means that the mechanical energy equals to the kinetic energy, since the equations are the same. Are the equations for potential energy correct?

 

[*Jessica*]

Or do these potential energies cancel out each other and all we are left with is 1/2(mv^2)?

 

[Sunil Simha]

Find U₁-U₂ and see for yourself.

So then,what would be the final kinetic energy?

 

[*Jessica*]

1/2(mv^2). Thank you!
Can you help with part 3, please?

 

[Sunil Simha]

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?

 

[*Jessica*]

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?

 

[Sunil Simha]

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.

 

[*Jessica*]

K2=0.5(1.45e20)(2.0)^2 + (-(G(1.45e20)(6e24))/6.4e6)-(G(1.45e20)(6e24))/3.84e8)

 

[Sunil Simha]

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 U_moon-U_ground?

 

[*Jessica*]

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.

 

[Sunil Simha]

The answer to part 2 seems correct
Hope you understood what to do with part 3

 

[*Jessica*]

Not really.. I don't understand the 60% part. It's 60% of what energy? The kinetic one I calculate in part 1?

 

[Sunil Simha]

No, I think they mean 60% of the nuclear blast's energy.

 

[sankalpmittal]

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 ?)