Would a standard howitzer on the Moon’s surface be able to hit Earth?

According to some numbers I found, the escape velocity of the moon is about 2.4km/s, plus about 800m/s for the moon-earth transfer orbit.
Artillery on the high end is about 1000m/s.
So if the gun was already in Lunar orbit, you could fire a shell to Earth. If it is on the surface of the moon, then no.

Not a chance, Even if you could get lift from the moon, the shell is far too small to survive getting through earth’s atmosphere. It will burn up before it ever gets through.

Remember John Glenn and the heat shield issue? It turned red hot during reentry, and it was made for heat.

A howitzer shell – impossible.

As an undergrad, I remember a similar homework problem. Can a person hit a
golf ball out of the moons’ gravitational pull… probably not. A howitzer most
definitely could blast a shell out of the gravitational pull of the moon.
Getting the shell to earth requires a lot of math and a bit of luck.

So I guess I would say yes!

@elbanditoroso
Re-entry vehicles are designed to offer as much air resistance as possible, to maximize their aerobraking capability.
Also, most of the heat generated is not due to friction, but due to pressure acting upon the flat surface of the vehicle.
Artillery shells are gyrostabilised aerodynamic projectiles, so if they enter the atmosphere tip first, they would probably make it to the ground just fine.

[Mod Says] This question is in the General Section. All responses must be helpful and on-topic. Furthermore, answers like “just Google it” and “look it up yourself” are always considered unhelpful and/or off-topic.

I came to this question late and I was like, “It’s a straightforward interesting question that I’ve never really thought about. What the heck is there to moderate?

Anyway, I disagree with @ragingloli on the reentry idea. I’m thinking that even though they are designed to be aerodynamic by the time the shells would get to Earth from the Moon and in reentry they will start to tumble and definitely free fall as there is no acceleration on them. They’d burn up.

Rocketguy would know the answer definitively as he’s a satellite engineer.

That can’t be right. The shell is designed to be ballistically suited to traveling through the air. And gravity forces acceleration, the reason things burn up to begin with. And for @elbanditoroso, the trick isn’t to worry about propulsion to the earth. If the shell is accelerated to the moon’s escape velocity, if aimed properly, the earth’s gravity will handle the rest.

And if explosives were eliminated from the shell and a solid core substituted the survivability of it to the ground might be arranged as surely as it would the warhead on an ICBM.

Actually, if a shell did make it to Earth, it would slow until only reaching a maximum falling speed of 176 feet per second, not fast enough to burn up. The shell didn’t begin it’s re-entry at the same speed as meteors, which do, generally, burn up.

Again, why do you think those heat shields are on the shuttles? Come on, you guys KNOW all this stuff. You,‘ve LIVED THROUGH IT.

Looking at my answer it I realize was imprecise, a bit lazy, and factually incorrect. When I said they would not have acceleration on them what I meant is that they do not have a rocket attached to them providing the projectile with a force—it is ballistic. They do, of course feel negative acceleration as they enter the atmosphere. That’s where the friction comes in and why they would burn up. (And why reentry craft need to have a heat shield as @hp said)

A standard howitzer fires a projectile with only a third of the moon’s escape velocity and so it will come crashing back to the moon at around 800m/s. It would never reach the Earth or the Earth’s atmosphere.

A re-entry of a space craft is different because it was previously orbiting at about 5 mps +, much faster than an incoming artillery shell would be going which would be less than 1 mps. (Modern tank rounds for Russian and US M1A1 120mm tank guns travel at just over 1mps.) The space craft is coming in at an angle, not straight in like I assume the artillery shell would.

The re-entry speed of the Apollo 11 mission was about 11km/s, so only about 20% faster than a hypersonic missile. Which makes sense, because as an object falls down towards Earth, it will accelerate. It only slows down, once it starts interacting with the atmosphere.
A projectile shot from the Moon towards Earth, assuming it is imbued with the necessary velocity, will probably have a similar speed when it hits the atmosphere.
The shell will not keep the same speed it is launched at.
First, it will decelerate as it moves away from the Moon, since the latter’s gravity slows it down. But once Earth’s gravity overcomes the Moon’s, it will accelerate.

@HP
The shuttle, and other vehicles, have the heat shield, because they were designed to offer as much resistance to the air as possible, and the resulting heat seen as a necessary consequence. They needed to do that, because the craft needed to slow down enough so it did not crash into the ground at meteoric speeds. Projectiles like bullets and shells are designed the opposite way, to minimise air resistance, so they very likely will not need heat shielding.

I agree 100%. That heat shield comment was to illustrate that ANY object drawn to the earth from space by the planet’s gravity will be accelerated to a velocity sufficient to generate extreme temperatures when passing through the planet’s atmosphere.

Remember also that the further an object is from Earth, the less gravitational pull there is. (The speed of gravity here on Earth is, basically, 32.1740 feet per second per second, which is the Standard Acceleration of Gravity, or “g.”)

The speed of gravity is 3×10^8m/s for the record. Acceleration due to gravity on earth is a paltry 9.8m/s

@Blackwater_Park There you go getting all relativistic on us.

To simplify the question, you wouldn’t need to fire a shell from the moon. You can do it from Earth’s orbit directly towards the surface. And in thinking about it, I’ve read enough science fiction to know that kinetic orbital bombardment is a thing so I’m changing my answer to “it would hit the surface and cause a lot of damage”

That “paltry” acceleration means the object’s speed is increasing by 9.8 meters per second EVERY second. So consider the kinetic energy imparted to an object the size of a marble accelerating at that rate for let’s say 10 minutes.

@Blackwater Park You don’t mean speed of gravity, you mean gravitational waves propagate at the speed of light.

As @ragingloli initially states, if the howitzer was already in lunar orbit it could fire a shot that would reach Earth. By the time it got there, Earth’s gravity would have accelerated it to around 11 km/s. That’s well over Mach 30. Most howitzer shells are made of steel, which melts at around 1400°C. Reentry would likely heat it up well above that, esp. at its pointy tip. Reentry capsules are normally blunt so that the super hot shock zone is as far as possible from the vehicle. In this case, the hot zone would be right at the tip, so a normal howitzer shell would never survive reentry. In fact, the explosive charge in normal shells would have gone off well before that point.

There are orbital bombardment concepts that use tungsten rods dropped from low Earth orbit: Rods from God. These would come down at “only” 8 km/s, and not melt because they would have a melting point of 3400°C. All kinds of fun would ensue at the landing sites.

@RocketGuy

“There are orbital bombardment concepts that use tungsten rods dropped from low Earth orbit: Rods from God.”

I remember reading about Project Thor:

“it provides better penetration then blast fragmentation type warheads. A 6.1 meter × 0.3 meter tungsten cylinder impacting at Mach 10 has a kinetic energy equivalent to approximately 11.5 tons of TNT (or 7.2 tons of dynamite).”

This has become interesting. First: what is Vietnam style howitzer? Looked it up
still unclear.

2. Can a howitzer catapult a shell (big bullet) beyond the pull of the moons’
gravity… the Lagrangian point for the moon.

3. If yes then the shell will still be within the Lagrangian point of the earth. It
will a satellite of the earth and by definition fall towards the earth. This transit
might take hundreds or even thousands of years (if not diverted) but definitely
will make it to earths’ lower orbit.

4. After this impossibly long journey will this shell finally get to rest on the
surface of the earth? Not a chance.

https://www.youtube.com/watch?v=OV1z7srUYBM

@HP “you mean gravitational waves propagate at the speed of light.” A.K.A. The speed of gravity. If the sun vanished instantly the earth would still act like it’s orbiting it for like 8 minutes.

@HP “That “paltry” acceleration means the object’s speed is increasing by 9.8 meters per second EVERY second. So consider the kinetic energy imparted to an object the size of a marble accelerating at that rate for let’s say 10 minutes”

There is a terminal velocity for such things because of drag. A 1.5 cm marble will reach say ~20m/s and no more.

If the object is being drawn towards Earth from lunar orbit 380,000+ km away, it would have plenty of time to accelerate in vacuum. Terminal velocity will not apply until about 150 km from Earth’s surface. That’s why Apollo and the upcoming Artemis capsules will hit Earth’s atmosphere at 11 km/s.

@Blackwater Park That terminal velocity for an object varies enormously with its shape, but here’s what really matters, the coefficient of drag will indeed dictate a terminal velocity, but at the point the object begins to interact with the atmosphere, it’s velocity is enormous and climbing. That trip through the atmosphere is relatively a blink of the eye. I too will cede the discussion to the @RocketGuy.

@RocketGuy, Before the illusion that kinetic energy is irrelelevant to this discussion passes, I would appreciate your take on the matter in regard to those tungsten rods and the destruction implicit due to kinetic energy.

Equivalent to the MOAB bomb at about 11 tons TNT:
https://aviationweek.com/defense-space/how-does-mother-all-bombs-stack

@RocketGuy Um…(Apollo came back and) Artemis will come back, orbit(ed) the Earth, then re-enter(ed) the atmosphere…
It’s not straight in.
If I remember correctly, when Apollo space crafts left for the moon, they attained 7 mps to escape their Earth orbit. They can’t go faster than that and re-enter. (I didn’t calculate what 11kps was compared to 7mps.)

@ragingloli It’s worth remembering that escape velocities are calculated assuming no other gravitational effects. An artillery shell fired towards Earth from the Moon will come under the influence of Earth’s gravitational pull. Escape velocities are meaningless in that event.

@kritiper : 7 m/s = 11 km/s. Whether they aim towards the center of Earth or the edge of the atmosphere, speed will be pretty much the same – it’s a 4000 mile deviation from a 250,000 mile trip. And you’re right about reentry speed. Orion is designed for 7 m/s reentry, SpaceX Dragon is designed for only 5 m/s.