Driving on the moon

Its a thousand years from now, there is a freeway round the moons' equator and it has been surveyed and evened out so it's perfectly circular within a mm. So this electric car with a normal drive like an electric car of today, but the wheels can turn 50,000 RPM if they have to or more, they are a meter in diameter and can take the spinning forces involved without flying apart. The question is, can you drive it into some sort of orbit, even if its only a cm above the ground?

Originally posted by sonhouse
Its a thousand years from now, there is a freeway round the moons' equator and it has been surveyed and evened out so it's perfectly circular within a mm. So this electric car with a normal drive like an electric car of today, but the wheels can turn 50,000 RPM if they have to or more, they are a meter in diameter and can take the spinning forces involved w ...[text shortened]... question is, can you drive it into some sort of orbit, even if its only a cm above the ground?

A very interesting idea!

If you take orbial energy into account I intuitively don't think so.

To go into a circular orbit, one has to give two push impulses. One to leave the ground, one to convert the elliptic orbit into a circular one. If you can do one, you can't do the other because your wheels don't touch the road anymore and no push is possible, if you don't have rocket drive.

So I say no - but I can't explain it further than this.

Originally posted by sonhouse
Its a thousand years from now, there is a freeway round the moons' equator and it has been surveyed and evened out so it's perfectly circular within a mm. So this electric car with a normal drive like an electric car of today, but the wheels can turn 50,000 RPM if they have to or more, they are a meter in diameter and can take the spinning forces involved w ...[text shortened]... question is, can you drive it into some sort of orbit, even if its only a cm above the ground?

I think you can. Given the information available, the top speed of the car is:

50,000 RPM * (1 x pi) m/rev * (1/60) min/s = 2618 m/s

Now, the gravity of the moon is about 1.62 m/s^2, but the centripetal acceleration required to keep the car on the road is:

a = v^2/R
= (2618)^2 / 1738000
= 3.94 m/s^2

This is greater than the gravity supplied by the moon, so eventually the car would start driving in a straight line and shoot into orbit. Just for fun, the minimum top speed required is:

v(req) = SQRT(R*a)
= SQRT(1738000*1.62)
= 1678 m/s

Taking a quick gander at Wikipedia, the speed of sound on Earth is approximately 344 m/s. Of course the moon has little to no atmosphere, so this wouldn't be a problem, but that's one fast car!

Originally posted by PBE6
I think you can. Given the information available, the top speed of the car is:

50,000 RPM * (1 x pi) m/rev * (1/60) min/s = 2618 m/s

Now, the gravity of the moon is about 1.62 m/s^2, but the centripetal acceleration required to keep the car on the road is:

a = v^2/R
= (2618)^2 / 1738000
= 3.94 m/s^2

This is greater than the gravity supplied by th ...[text shortened]... the moon has little to no atmosphere, so this wouldn't be a problem, but that's one fast car!

But you can't get into a circular orbit one cm above the road around one full revolution. It's only a matter of time before you bounce back to the road again. You cannot stay in orbit, becuse it cannot be perfectly circular in this scenario.

Originally posted by PBE6
I think you can. Given the information available, the top speed of the car is:

50,000 RPM * (1 x pi) m/rev * (1/60) min/s = 2618 m/s

However, this assumes that the max RPM of the wheels can be converted into straight-line speed. As the car approaches orbital speed the force it exerts on the ground will decrease, and so you are likely to reach a point where their isn't enough grip to accelerate it further.

(I don't think this argument doesn't prove it isn't possible, but it shows that the calculation needs to be a bit more complicated).

Originally posted by FabianFnas
To go into a circular orbit, one has to give two push impulses. One to leave the ground, one to convert the elliptic orbit into a circular one. If you can do one, you can't do the other because your wheels don't touch the road anymore and no push is possible, if you don't have rocket drive.

He didn't say it had to be a circular orbit.

I think you're right in that any orbit achieved via a pure straight line push would mean you just touch down twice per orbit. But the force needed to change this orbit by a couple of mm wouldn't be much.

Originally posted by mtthw
He didn't say it had to be a circular orbit.

Sonhouse: "The question is, can you drive it into some sort of orbit, even if its only a cm above the ground?"

Let's define the concept of orbit: I say that a orbit is off ground all the way round. If it is 1 cm off ground all the way round it is circular, if not it is elliptic. I think we can rule out a parabolic oand hyperbolic orbit here, because it is a closed orbit. If it tuoches the ground anyware round the moon, it is not an orbit. Do we agree on this?

If I say the orbit must be circular I am wrong, it doesn't matter if it is elliptic. But the main question stands: Can the vehicle leave the ground?

I say that when the vehicle has the speed it can leave the ground it cannot gain more speed so it actually can leave the ground becous the wheel lose its contact and friction with the ground. That's why it cannot leave the road. (The road has no bumps.)

Once I read a book about astronautics and from that got some insights about planetary (and lunary) orbits...

Originally posted by FabianFnas
Sonhouse: "The question is, can you drive it into some sort of orbit, even if its only a cm above the ground?"

Let's define the concept of orbit: I say that a orbit is off ground all the way round. If it is 1 cm off ground all the way round it is circular, if not it is elliptic. I think we can rule out a parabolic oand hyperbolic orbit here, because it k about astronautics and from that got some insights about planetary (and lunary) orbits...

One thing I was thinking, obviously you can get to a speed where you would have at least a sub-orbital velocity so you would be above the ground for a certain amount of time, coming down X amount of Km away, say 1000 Km just for grins. I am thinking if you are ready, the force that lifted you up would be exactly the same force you would have coming down and touching the road. I am thinking further that if this happened and you timed it right, like getting your motors revved just as you touched down, you would get another burst of speed, but would it be enough?
One thing that suprised the nuts off me in regular drag racing, is the fact that those drag racers regularly achieve 3 or more G's of acel.
So if you had similar tires and roadway on the moon, then it would seem to follow you could get more than one lunar G of acel, so when you touch down again, maybe you can get a lot more than one lunar G and that might make the differance.

Originally posted by FabianFnas
Sonhouse: "The question is, can you drive it into some sort of orbit, even if its only a cm above the ground?"

Let's define the concept of orbit: I say that a orbit is off ground all the way round. If it is 1 cm off ground all the way round it is circular, if not it is elliptic. I think we can rule out a parabolic oand hyperbolic orbit here, because it ...[text shortened]... k about astronautics and from that got some insights about planetary (and lunary) orbits...

Good point, gravity gives the car traction. At the point where the centripetal acceleration required and gravity balance, the car would have no traction and would not be able to overcome gravity. There would need to be a iump or some other irregularity on the track.

Originally posted by PBE6
[b]Good point, gravity gives the car traction. At the point where the centripetal acceleration required and gravity balance, the car would have no traction and would not be able to overcome gravity. There would need to be a iump or some other irregularity on the track.[/b

Our threads posted together. The point I am making is when it gets up enough velocity to get into a sub-orbtal, it would be the equivalent of hitting a bump when the orbit intersects the road again. There is one kicker that messes up our neat equaations thouigh: Mascons.
The moon is not prerfectly homogenous in density and the orbits of special lunar probes have been monitored to find those mascons (mass concentrations) under the surface. That would play havoc with our little thought experiment.

This is a very interesting problem! Math is a good part of the problem but reasoning is a more important part.

The bump is such an idea. If it works (I'm not soure it will be enough), you can have a lift off but with a sub orbital speed so you land again. During this contact you acelerate again to have another lift off, now with a higher orbit. But you will never have a lift off big enoug so you can be off ground all the waya round the moon. You need to impulses: One at the ground and one in the apogeum point of the orbit. From where do you get the extra thrust? So you fall down again.

Another complication: When you are off ground and you accelerate your tires your vehicle start to rotate in the counter rotation relative to your tires. The rotation momentum has to be the same so if you accelerate the tires you conter rotate the rest of the vehicle. All in good Newton laws about the action-reaction principle.

I still don't think it is possible to reach a lunar orbit, elliptic nor circular.

Originally posted by mtthw
[b As the car approaches orbital speed the force it exerts on the ground will decrease, and so you are likely to reach a point where their isn't enough grip to accelerate it further.

b]

I agree.

In order to maintain an orbit you need to accelerate the car faster than simply acheiving lift off the ground. Otherwise, gravity will immediately begin to pull the car back down to the ground.

It's impossible to make the car go faster once it's lifted off the ground.

Originally posted by FabianFnas
[b]Another complication: When you are off ground and you accelerate your tires your vehicle start to rotate in the counter rotation relative to your tires. The rotation momentum has to be the same so if you accelerate the tires you conter rotate the rest of the vehicle. All in good Newton laws about the action-reaction principle.b]

Ok, how bout if you immediately threw the gear shifter into reverse after you lift off. The tires would be rotating backwards, thus propelling the car forward if i understand your point.

Would this not provide you with additional forward momentum?

What...ever!

Originally posted by Alethia
What...ever!

That's your big contribution to this poser?