How fast to shoot a bullet for it to hit the moon?

[Treva31]

Ignoring the fact that the atmospheric friction would probably disintegrate it.

I know that to achieve LEO takes around 10km/s delta v.
But only 2km/s of that is to overcome the drag and gravity.
So you could shoot at 2km/s, get to LEO altitude then it falls back down because it is not at orbital velocity or trajectory right?

And I think that you need around 16km/s (10 to LEO plus 6 from LEO to moon) for a rocket to safely land on the moon from Earth.
But is it less than 16 if you just want to shoot a bullet and hit the moon?
Would it be 2 + 6 = 8km/s ??

Wouldn't it at least be less than the 11.2km/s escape velocity of the Earth since if you shot a bullet to the altitude of the moon (but not near the moon) it would eventually fall back to Earth, so you haven't escaped the Earths gravity.

 

[MarcusAgrippa]

Ignoring the fact that the atmospheric friction would probably disintegrate it.

I know that to achieve LEO takes around 10km/s delta v.
But only 2km/s of that is to overcome the drag and gravity.
So you could shoot at 2km/s, get to LEO altitude then it falls back down because it is not at orbital velocity or trajectory right?

And I think that you need around 16km/s (10 to LEO plus 6 from LEO to moon) for a rocket to safely land on the moon from Earth.
But is it less than 16 if you just want to shoot a bullet and hit the moon?
Would it be 2 + 6 = 8km/s ??

Wouldn't it at least be less than the 11.2km/s escape velocity of the Earth since if you shot a bullet to the altitude of the moon (but not near the moon) it would eventually fall back to Earth, so you haven't escaped the Earths gravity.

Use the conservation of energy to work this out. To reach the moon, the velocity of the bullet must never reverse. To reverse, it must pass through the value zero. Take the potential energy of the Earth and of the Moon into account. In the limiting case, the velocity will be zero at the first Lagrangian point. To push it over the Lagrangian point, its velocity needs to be infinitesimally more than zero there.

 

[DrStupid]

Take the potential energy of the Earth and of the Moon into account. In the limiting case, the velocity will be zero at the first Lagrangian point. To push it over the Lagrangian point, its velocity needs to be infinitesimally more than zero there.

That's not sufficient because the Moon orbits Earth with 1 km/s. It will run away before the bullet gets a chance to reach this speed. In a numerical simulation I did not reach the Moon with a speed minimum of less than 220 m/s (in the inertial system). The corresponding initial speed was 11091 m/s.

 

[MarcusAgrippa]

That's not sufficient because the Moon orbits Earth with 1 km/s. It will run away before the bullet gets a chance to reach this speed. In a numerical simulation I did not reach the Moon with a speed minimum of less than 220 m/s (in the inertial system). The corresponding initial speed was 11091 m/s.

My apologies. I assumed this was a freshman problem. I guess I should call myself "Dr Stupid", right?

 

[Treva31]

That's not sufficient because the Moon orbits Earth with 1 km/s. It will run away before the bullet gets a chance to reach this speed. In a numerical simulation I did not reach the Moon with a speed minimum of less than 220 m/s (in the inertial system). The corresponding initial speed was 11091 m/s.

Cool thanks.
How do you do a numerical simulation? Is there some software i can download?

 

[DrStupid]

How do you do a numerical simulation?

I just used EXCEL and a velocity verlet algorithm.

 

[Treva31]

I just used EXCEL and a velocity verlet algorithm.

So just to clarify, the 11091 m/s doesn't include any velocity needed to overcome the friction of the atmosphere right? Just the gravity.

Any chance I could get a copy of the Excel file?

 

[tony873004]

Here's a numerical simulation. Change "Venus" to "bullet" if you want to edit the bullet's velocity or position. Then press Play, the > button on the "Time Step" interface.
http://orbitsimulator.com/gravitySimulatorCloud/simulations/1449002521127_bullet.html

 

[tony873004]

I just noticed this doesn't display properly unless you have a wide aspect ratio screen.
If you can't see the Moon, press "A" on your keyboard, then press "-" minus under the word Zoom to zoom out.

 

[Treva31]

I just noticed this doesn't display properly unless you have a wide aspect ratio screen.
If you can't see the Moon, press "A" on your keyboard, then press "-" minus under the word Zoom to zoom out.

Wow that is awesome!
Thanks so much guys.

 

[DrStupid]

So just to clarify, the 11091 m/s doesn't include any velocity needed to overcome the friction of the atmosphere right? Just the gravity.

Yes, it is the initial velocity in the inertial system (not the relative velocity in the rest system of Earth) for a free fall trajectory without any interactions except gravity.

Any chance I could get a copy of the Excel file?

Here it is: http://www.drstupid.de/temp/moonflight.xls

I played a little bit with the starting values and reduced the initial velocity to 11080 m/s.

Here is the same simulation using JavaScript and Runge-Kuta-Nyström: http://tinyurl.com/j5anj3m

 

[Treva31]

Yes, it is the initial velocity in the inertial system (not the relative velocity in the rest system of Earth) for a free fall trajectory without any interactions except gravity.
Here it is: http://www.drstupid.de/temp/moonflight.xls

I played a little bit with the starting values and reduced the initial velocity to 11080 m/s.

Here is the same simulation using JavaScript and Runge-Kuta-Nyström: http://tinyurl.com/j5anj3m

Perfect thank you!

 

[Janus]

Yes, it is the initial velocity in the inertial system (not the relative velocity in the rest system of Earth) for a free fall trajectory without any interactions except gravity.
Here it is: http://www.drstupid.de/temp/moonflight.xls

I played a little bit with the starting values and reduced the initial velocity to 11080 m/s.

/QUOTE]
That comes out close to my "quick and dirty" calculation method which gave me an answer of 11087 m/s

Step 1: Calculate the total energy of the bullet with respect to the Moon with the difference in velocity equal to the orbital velocity of the Moon.
Step 2. Use this to get the semi-major axis of the hyperbolic path the bullet takes wrt the Moon.
Step 3. Assuming a perilune of 1738 km ( so the path intersects the Moon), work out the eccentricity of said path.
Step 4. From the eccentricity work out the deflection angle of the path.
Step 5. From the deflection angle, compute the impact parameter
Step 6. Calculate the needed initial bullet velocity needed to get to an altitude equal to the moon's orbital radius minus the impact parameter.

 

[Treva31]

Is there any way to reduce the impact speed?

Or might it be possible shoot the bullet into lunar orbit?
A decaying/sub-orbit that ends with a softer landing would be good too!

Am I right in saying it is not possible to shoot anything into Earth orbit without a trajectory correction? Even if shot at a very shallow angle (1 degree) at a precise velocity?The mass of the "bullet" would be 10kg. Would a larger mass (100kg) help the above questions?

 

[Treva31]

Could you use the gravity of the moon to slow the bullet further?

If you shot just "in front" of the moons path, and the bullet moves around to the darkside, it is pulled around the moon and towards the surface by the moons gravity, slowing it down, before it finally lands, at less than 220 m/s?
I'm not sure how far around would be optimal, maybe down almost straight away (we aim closer to the surface), or maybe half an orbit or more.

See my crude illustration attached.

 

[Janus]

Could you use the gravity of the moon to slow the bullet further?

If you shot just "in front" of the moons path, and the bullet moves around to the darkside, it is pulled around the moon and towards the surface by the moons gravity, slowing it down, before it finally lands, at less than 220 m/s?
I'm not sure how far around would be optimal, maybe down almost straight away (we aim closer to the surface), or maybe half an orbit or more.

See my crude illustration attached.

The best you can do is place it at rest with respect to the Moon and then have it fall toward the Moon from there, but it will still impact at over 2300 m/sec.

 

[Treva31]

The best you can do is place it at rest with respect to the Moon and then have it fall toward the Moon from there, but it will still impact at over 2300 m/sec.

We already showed that it can hit directly at only 220 m/s without any gravity assist.

 

[russ_watters]

We already showed that it can hit directly at only 220 m/s without any gravity assist.

That's not what that post says: that post was referring to the speed at the Legrange point.

There really isn't anything that can be done without power to slow the impact speed: it can't be much below escape velocity otherwise.

Am I right in saying it is not possible to shoot anything into Earth orbit without a trajectory correction?

Correct.

 

[tony873004]

Janus is right. You can't approach the Moon from beyond its orbit and crash into it with anything less than its escape velocity. It violates the Law of Conservation of Energy. At best you can have 0 Potential Energy at infinity. Subtract that from its PE at the surface (-GMm/r) for a change in Potential Energy. All that lost PE will become Kinetic Energy. So about 2300-2400 m/s is the slowest you can land unpowered.

 

[Treva31]

That's not what that post says: that post was referring to the speed at the Legrange point.

Ah OK, sorry I misunderstood.

 

[russ_watters]

Janus is right. You can't approach the Moon from beyond its orbit and crash into it with anything less than its escape velocity. It violates the Law of Conservation of Energy. At best you can have 0 Potential Energy at infinity. Subtract that from its PE at the surface (-GMm/r) for a change in Potential Energy. All that lost PE will become Kinetic Energy. So about 2300-2400 m/s is the slowest you can land unpowered.

Just to head off a possible nitpick by others, we had a thread a year or so ago regarding the speed of comet impacts. I worded my statement similarly and "lost" the argument over it: it turns out that because of the multiple objects and motions involved, when "catching-up" to the orbit of another object, you can impact at a *bit* less than escape velocity. If I remember correctly, it's within a few percent.

Here's that thread:
https://www.physicsforums.com/threads/meteorite-crash-in-russia.671921/page-3