Muzzle velocity

[Gareth56]

If you could fire a rifle on the Moon (i assume you can) what would the muzzle velocity of a rifle need to be to send a bullet into orbit around the Moon?

[YT2095]

2.4 Km/Sec. or less depending on the Type of Orbit you want.

[swansont]

If you stand on the surface and shoot parallel, 1.68 km/s should do. Step out of the way after firing.

[Gareth56]

I got 1.7km/s by equating Fgrav to Fcent then solving for v. Is that how you did it?

[YT2095]

who? me?

 

No, I just read the Moons Escape velocity out of a book and the rest is derived from that

[MrMongoose]

who? me?

 

No, I just read the Moons Escape velocity out of a book and the rest is derived from that

 

How exactly do you derive it from escape velocity? I need to find myself a course in orbital mechanics if I'm ever going to get that job as a rocket scientist:mad:

[YT2095]

well, as i posted, the Escape velocity according to my book is 2.4Km/Sec, meaning that if you fire that projectile upwards perfectly perpendicular to the force of gravity, it will eventually be Free of the moons pull or ability to drag it back down again, ergo anything Less than 90 degrees or a lower velocity, will attain Some sort of orbit

[theCPE]

meaning that if you fire that projectile upwards perfectly perpendicular to the force of gravity

 

Isn't upwards parallel to gravity?

 

You fire the projectile parallel to the surface, perpendicular to the center of gravity.

 

ergo anything Less than 90 degrees or a lower velocity, will attain Some sort of orbit

 

You can put the projectile in orbit with a parallel trajectory to the surface.

 

Escape velocity just requires you to travel parallel to the surface fast enough that as the moon/planet gravity pulls you back to the surface you keep "missing" it.

[Gareth56]

well, as i posted, the Escape velocity according to my book is 2.4Km/Sec, meaning that if you fire that projectile upwards perfectly perpendicular to the force of gravity, it will eventually be Free of the moons pull or ability to drag it back down again, ergo anything Less than 90 degrees or a lower velocity, will attain Some sort of orbit

 

But if you were firing the bullet horizontally (and assuming nothing would impede its path) then the escape velocity doesn't enter into the debate and only Moon's acceleration due to gravity and radius are relevant.

[theCPE]

But if you were firing the bullet horizontally (and assuming nothing would impede its path) then the escape velocity doesn't enter into the debate and only Moon's acceleration due to gravity and radius are relevant.

 

Thats what escape velocity refers to. Speed required for the projectile to leave whatever planet/object without any force, if you fire a projectile horizontally escape velocity is the concept to determine the projectile's fate.

[Gareth56]

Thats what escape velocity refers to. Speed required for the projectile to leave whatever planet/object without any force, if you fire a projectile horizontally escape velocity is the concept to determine the projectile's fate.

 

But the trick is to determine what velocity just prevents the bullet from flying off into space and maintains the bullet in a circular orbit. How this can be achieved using the escape velocity figure and something else escapes me.

[MrMongoose]

well, as i posted, the Escape velocity according to my book is 2.4Km/Sec, meaning that if you fire that projectile upwards perfectly perpendicular to the force of gravity, it will eventually be Free of the moons pull or ability to drag it back down again, ergo anything Less than 90 degrees or a lower velocity, will attain Some sort of orbit

 

I can't see where you posted that. You posted the escape velocity, with no derivation of the velocity needed for orbit following.

 

The OP's question asked what velocity is required to cause orbit, there is clearly going to need to be a "vertical" component to overcome gravity to such an extent that when the bullet stops moving radially, a further "horizontal" component will give it the correct speed to orbit under the effects of centripetal acceleration at that radius.

 

Is 1.7 therefore the minimum speed possible courtesy of pythagoras?

 

If thats not how swansont and gareth did it, I'll see what my method gives when I get home.

[theCPE]

But the trick is to determine what velocity just prevents the bullet from flying off into space and maintains the bullet in a circular orbit. How this can be achieved using the escape velocity figure and something else escapes me.

 

Yes, orbital velocity and escape velocity are different.

 

I was making clarifications about escape velocity because there were some misconceptions going on.

 

But its a similar idea. You find the velocity at which the projectile travels parallel to the surface enough per unit time to completely negate the distance it is pulled toward the surface based on the gravitational constant etc.

 

Orbit velocities depend of course on the orbit altitude as well.

 

Fore specific examples and equations try googling the two.

[Gareth56]

<snip)You find the velocity at which the projectile travels parallel to the surface enough per unit time to completely negate the distance it is pulled toward the surface based on the gravitational constant etc.

<snip>

QUOTE]

 

Yes and you don't need to know anything about what the escape velocity is you just need the radius of the planet and its acceleration due to gravity value. Hence the 1.7km/s speed for the Moon.

[theCPE]

Of course not, didn't say you did.

 

The way you calculate the two are similar. All my prior posts were aimed at clarify some misconceptions or whatever someone else posted about escape velocity, hence my focus on escape velocity.

 

[swansont]

I got 1.7km/s by equating Fgrav to Fcent then solving for v. Is that how you did it?

 

Yes.

[MrMongoose]

Can someone tell me where I'm going wrong here?

 

Parallel to the surface of the moon there will be no acceleration, so the initial horizontal component of velocity will be equal to the final horizontal component.

 

The final component is given by equating the necessary centripetal force to the gravitational force, i.e

 

[math]

m\frac{v^2_h}{R}=\frac{mMG}{R^2}

[/math],

 

which is

 

[math]

v^2_h=\frac{MG}{R}

[/math].

 

(M is the moon's mass, and m that of the bullet. v is velocity with subscript h being the horizontal component and G is the gravitational constant. R is the radius of the orbit).

 

Letting subscript v denote the vertical component and r be the radius of the moon, conservation of energy can be applied-

 

[math]

m\frac{v^2_h}{2} + m\frac{v^2_v}{2} + \frac{mMG}{r}=\frac{mMG}{R} + m\frac{v^2_h}{2}

[/math],

 

or,

 

[math]

v^2_v = 2MG(\frac{1}{R}- \frac{1}{r})

[/math].

 

Now, the muzzle speed of the bullet is simply given by

 

[math]

v^2 = v^2_v + v^2_h= 2MG(\frac{1}{R}- \frac{1}{r}) + \frac{MG}{R}

[/math]

 

i.e.

 

[math]

v^2 = 2MG(\frac{3}{2R}- \frac{1}{r})

[/math]

 

Which implies that R=3r/2 v=0, i.e. If im sitting here at rest in my chair now, that should be enough velocity to put me into orbit at 1.5x the Earth's radius.

[Gareth56]

Of course not, didn't say you did.

 

The way you calculate the two are similar. All my prior posts were aimed at clarify some misconceptions or whatever someone else posted about escape velocity, hence my focus on escape velocity.

 

 

In that case I'd be very interested in seeing the method using the escape velocity figure.

 

Can someone tell me where I'm going wrong here?

 

 

Yes, you're using the wrong equations. You only need the equate the gravitational force (mg) to the centripetal force (mv^2/r) the solve for V.

[MrMongoose]

Fixed it now- any ideas?

[Sisyphus]

I'd just like to point out that ANY velocity that's less than escape velocity can be orbital velocity, if you don't specify anything else. I have noticed, however, that most people get annoyed when their orbits intersect the surface. Will that be a problem for you, as well?

[MrMongoose]

Ah, so thats the velocity for an orbit along the surface! Well a few hundred metres below the surface, but im assuming thats because I worked to 3 s.f

 

Cheers

 

You only need the equate the gravitational force (mg) to the centripetal force (mv^2/r) the solve for V.

 

I can see where you get mg from now but how did you know the slowest muzzle velocity would correspond to orbit at the surface?

[Gareth56]

Ah, so thats the velocity for an orbit along the surface! Well a few hundred metres below the surface, but im assuming thats because I worked to 3 s.f

 

Cheers

 

 

 

I can see where you get mg from now but how did you know the slowest muzzle velocity would correspond to orbit at the surface?

 

Because you're balancing the weight of the bullet with the centripetal acceleration. So the mass doesn't enter into it.

[swansont]

Fixed it now- any ideas?

 

PE is negative, so you've got the wrong sign.

[thedarkshade]

If you could fire a rifle on the Moon (i assume you can) what would the muzzle velocity of a rifle need to be to send a bullet into orbit around the Moon?

You can find that out by using first cosmic velocity formula [math]v=\sqrt{gR}[/math]

[swansont]

You can find that out by using first cosmic velocity formula [math]v=\sqrt{gR}[/math]

 

Which is what you get when you set the gravitational and centripetal force terms equal, which has already been discussed. (since g = Gm/R²)