What Is the Minimum Velocity Factor for a Lunar Orbit?

[Physgeek64]

Homework Statement

Compute the escape velocity for a projectile fired from the surface of the moon.
The moon’s radius is 1700 km, and its surface gravity is $1.62 m s^(−2)$ . Neglect the effect
of the Earth. A second projectile is fired horizontally from the lunar surface with a velocity f times the escape velocity. Find the minimum value of f for the projectile to
go into orbit.

Homework Equations

The Attempt at a Solution

So I have done the first part by simply equating the energy to zero $1/2*m*v^2 = GMm/R$ and noting that $g=GM/R$ its easy to show that $v=sqrt(2*g*R)$

However, I am confused about the second part since I have not been given a radius of orbit. So my first thought was that the minimum energy of an orbit is a circular orbit, so am I supposed to find the energy required to put the projectile into an orbit of radius R? $1/2*m*(fv)^2=GMm/R$ ?

Many thanks :)

 

[BvU]

the minimum energy of an orbit is a circular orbit

Maybe, maybe not. Can you underpin this ? And: if so, is there going to be a transition from the original orbit to this circular orbit ? How can that be organized physically ?

PS Note that I'm not saying your assumption is incorrect...just that you may want to imagine how this minimum speed case comes about; what is (are) the condition(s)

 

[Physgeek64]

Maybe, maybe not. Can you underpin this ? And: if so, is there going to be a transition from the original orbit to this circular orbit ? How can that be organized physically ?

PS Note that I'm not saying your assumption is incorrect...just that you may want to imagine how this minimum speed case comes about; what is (are) the condition(s)

Well I would suppose that it is the minimum energy since this is when the effective potential its minimum, and is the only orbit that has no component to its radial velocity, whereas every other orbit (which already has a greater $U_e$) would also gain more energy due to this radial component. Alternatively you can differentiate the orbital equation with respect to r, set the derivative to zero and find that the centripetal force is equal to the gravitational force, and can hence infer circular motion :)
Im not sure- the question wasn't very clear. My problem was that orbits are always closed, so it can't be elliptical since this would cause the projectile to crash back into Earth from its original position. The only other possibility I could think of would be to have a circular orbit in which the projectile is essentially orbiting just above the surface of the moon?

Thank you :)

 

[Simon Bridge]

Think about a projectile fired horizontally at a very low energy, say 1 or 2m/s ... what path does it follow? Where does it end up? Is this an "orbit"?
How about 5 or 6m/s? Where does the projectile end up compared with the first case? Is this an orbit?

Where would the projectile end up for the trajectory to be considered an orbit?

[edit]... you may have done this from a different perspective.

 

[Physgeek64]

Think about a projectile fired horizontally at a very low energy, say 1 or 2m/s ... what path does it follow? Where does it end up? Is this an "orbit"?
How about 5 or 6m/s? Where does the projectile end up compared with the first case? Is this an orbit?

Where would the projectile end up for the trajectory to be considered an orbit?

[edit]... you may have done this from a different perspective.

I would assume it would be the limit for which the projectile just makes it around the earth- i.e. performs circular motion at a radius ever so slightly larger than that of the moon, as I had said. Sorry if I'm missing something obvious :P Many thanks :)

 

[BvU]

The only other possibility I could think of would be to have a circular orbit in which the projectile is essentially orbiting just above the surface of the moon?

Sounds reasonable to me: the condition being it doesn't fall faster than at the rate 'the ground curves down'. So, did you manage to convince yourself and can we strike the question mark ?