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

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Discussion Overview

The discussion revolves around the theoretical question of how fast a bullet would need to be shot in order to hit the Moon, considering various factors such as atmospheric friction, gravitational influences, and orbital mechanics. Participants explore the implications of different velocities and trajectories, as well as the feasibility of achieving such a task with a bullet.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants suggest that achieving Low Earth Orbit (LEO) requires a delta v of around 10 km/s, with only 2 km/s needed to overcome atmospheric drag and gravity.
  • There is a proposal that to reach the Moon, a bullet might need a speed of at least 8 km/s, combining the 2 km/s to reach LEO and an additional 6 km/s to travel from LEO to the Moon.
  • Others argue that the bullet would need to exceed the escape velocity of 11.2 km/s to avoid falling back to Earth, although some suggest it might be less than this if the bullet does not escape Earth's gravity.
  • A participant mentions using conservation of energy to analyze the problem, indicating that the bullet's velocity must remain positive to reach the Moon.
  • Another participant shares results from a numerical simulation indicating that a minimum speed of 11091 m/s was required to reach the Moon, with an initial speed of 220 m/s in the inertial system.
  • There are discussions about the effects of atmospheric friction and the initial conditions necessary for the bullet's trajectory.
  • Some participants inquire about the methods used for numerical simulations, sharing tools like Excel and JavaScript for modeling the bullet's path.
  • There are suggestions about potentially using the Moon's gravity to slow the bullet for a softer landing or to achieve a lunar orbit.
  • Questions arise regarding the feasibility of shooting a bullet into Earth orbit without trajectory corrections, and whether a larger mass would influence the outcome.

Areas of Agreement / Disagreement

Participants express a range of views, with no consensus on the exact speed required for the bullet to hit the Moon. There are competing models and calculations presented, and the discussion remains unresolved regarding the optimal conditions and parameters for achieving the goal.

Contextual Notes

Limitations include assumptions about atmospheric conditions, the effects of gravity, and the simplifications made in the numerical simulations. The discussion also highlights the complexity of orbital mechanics and the need for precise calculations.

Who May Find This Useful

This discussion may be of interest to those studying physics, particularly in the areas of mechanics and orbital dynamics, as well as individuals curious about the practical implications of shooting projectiles at astronomical distances.

Treva31
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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.
 
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Treva31 said:
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.
 
MarcusAgrippa said:
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.
 
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DrStupid said:
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?
 
DrStupid said:
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?
 
Treva31 said:
How do you do a numerical simulation?

I just used EXCEL and a velocity verlet algorithm.
 
DrStupid said:
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? :wink:
 
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.
 
  • #10
tony873004 said:
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.
:smile:
 
  • #11
Treva31 said:
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.

Treva31 said:
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
 
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  • #12
DrStupid said:
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! :smile:
 
  • #13
DrStupid said:
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.
 
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  • #14
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?
 
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  • #15
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.
 

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  • #16
Treva31 said:
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.
 
  • #17
Janus said:
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.
 
  • #18
Treva31 said:
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.
 
  • #19
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.
 
  • #20
russ_watters said:
That's not what that post says: that post was referring to the speed at the Legrange point.

Ah OK, sorry I misunderstood.
 
  • #21
tony873004 said:
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
 
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