There are two identical uniform spherical planets of radius R. The first has its center at the origin of the xyz coordinate system. The second has its center at (2R, 0, 0). The planets are touching.
A projectile is launched from the "North Pole" of the first planet at (0, 0, R) with its initial velocity pointed in the direction of the vector (1, 0, 1).
Let the escape speed relative to the planet's surface be ve. Note that here, the escape escape is for a single planet in isolation (following the typical convention).
With the given launch vector, let v0 be the minimum launch speed for the projectile to reach the "North Pole" of the second planet at (2R, 0, R).
How are the two speeds ve and v0 related?
Assume each planet has a mass, M and radius R.
Then escape velocity for either is sqrt(2GM/R), with G being the gravitational constant. This is V(e)
But in reality the main influence on the projectile will be the barycentre of the system at the point where they touch on the equator. This system has a mass of 2M, and a radius of R*2^(1/3). The projectile at rest at the North Pole is already in orbit in this frame, at a height equal to around 1/8th of the radius of the larger system.
At the surface of the notional sphere, escape velocity would be sqrt(2G(2M)/(R*2^(1/3))). At the North Pole however, this reduces to sqrt(2G(2M)/(9R/8*2^(1/3))), that is (2 sqrt(2))/3 times the original value. The same as launching from a very high mountain, say.
Now using v(o)=v(e)/sqrt(2), we have (sqrt(2G(2M)/(9R/8*2^(1/3))))/sqrt(2) or (2*2^(1/3)*sqrt(G(2M)))/(3sqrt(R)), as the speed at which the (unpowered) projectile will be launched from the North Pole. This is V(o)
V(e)/V(o) = sqrt((2GM)/R)/(3*2^(5/6) sqrt(R) sqrt(G(2 M))), or 1/(6*2^(1/3) R)
(As the problem hints, G and M do not matter for this purpose.)
This is a maximum value as it does not rule out the potential for other more efficient orbits.
Edited on June 27, 2019, 7:23 am
|
|
Posted by broll
on 2019-06-21 22:50:44 |
Some thoughts and a maximum
