We loft something into space somewhere that’s short of escape velocity, but more than the what’s needed to orbit. Just a “higher” orbit?
There is no in-between.
If you (or any of your loved ones) are in circular orbit and get some acceleration you go into an elliptical orbit with higher apogee and same perigee but still “orbit”.
If the acceleration is sufficient then you escape and don’t return.
Either you return or you don’t. If you return you are in orbit. If you don’t return then you are not.
An orbit with a higher eccentricity (a bigger difference between apogee and perigee), and therefore an orbit that takes longer to complete. When you hit “escape velocity” the time to complete the orbit reaches infinity…
When the Apollo Astronauts fired their thrusters to go to the moon they didn’t actually hit escape velocity. They entered into a highly elliptical orbit around Earth with the apogee (the point furthest from Earth) near the moon.
If the astronauts had continued to add thrust they would have escaped Earth but then entered into orbit around the sun. If you burn long enough to escape the sun then you’d enter orbit around the milky way. If you burn past that you’d enter into intergalactic space.
I think the answer to the OP would be that you would return given in infinity time to do so, that would be the theoretical middle point…
Maybe not, but they surely got close to it. The surface of the earth is 4000 miles from the center, and the moon is 240,000 miles away, so the gravitational acceleration of earth at that distance is 4000[sup]2[/sup]/240000[sup]2[/sup] = 1/3600 what it is on the earth’s surface; I don’t have the equation in front of me, but I suspect there’s not much potential energy to be gained by moving farther from the earth than that.
Depending on where you look on the interwebz, Apollo’s reentry speed (after “falling” from the moon back to earth) appears to be around 24,200 MPH; escape velocity is 25,053 MPH, so yes, they were pretty close to it.
Thank you. I think this is what I was looking for.
Note that you need more than just orbital velocity to orbit. The velocity has to be pointed in the right direction, and has to start from a point that’s not on the planet.
For example, fire a projectile up at an orbital velocity (ignore air resistance). It’ll go into orbit, but the path of that orbit will include the starting point where it started, which is on the ground, so wham. To get into orbit, you have to first get off the planet (and above the atmosphere, to keep things simple) and then fire the engines to hit orbital velocity, in the direction you want your orbit to take (dramatically oversimplifying).
Note that all orbits / trajectories are conic curves. A normal orbit is going to be an ellipse, of which a circle is just a particular case. If you have “excessive” speed then the trajectory is an hyperbola and there is no return. A parabola is the limit between a parabolic and a hyperbolic trajectory.
If you are in a circular orbit and you fire your rockets tangentially to gain speed then you go into an elliptical orbit with higher apogee and same perigee. When you get to your new, higher, apogee you can again fire your rockets to gain speed so that you will not fall back to your previous perigee but will remain in higher orbit.
If you gain enough speed you will “climb” away indefinitely with hyperbolic trajectory.
It should be understood that the escape speed is not a single value for a celestial body, but is the instantaneous speed at any given distance from the body center of mass such that the characteristic energy (C3, the specific energy in excess of that required to achieve an escape trajectory in a two body problem) is greater than 0. Note that C3 is exactly twice the sum of kinetic and potential orbital energy divided by the standard gravitational parameter (specific orbital energy, ϵ) which means that as the periapsis or eccentricity of the orbit increases and the specific orbital energy reduces, the C3 decreases. At the point at which kinetic and potential energy are equal the C3 is zero, and the object is in a parabolic orbit by definition. Note that while orbit is commonly understood to refer to only elliptical orbits, any object within the gravitational sphere of influence of a body can be considered to be in orbit, be it a periodic orbit or a non-returning orbit on a parabolic or hyperbolic trajectory.
So, in a strictly two body universe, it is entirely possible for an object to be any arbitrary distance from a massive body and still remain in a (very slow, long period) elliptical orbit so long as the kinetic energy is low enough that the C3 is negative. In practice, a object in space is always subject to gravitational fields in beyond its primary body, and if the C3 is sufficiently close to 0, at some point another influence will perturb it into an escape trajectory, hence why low mass objects can enter non-conic section orbits such as figure-8, horseshoe, and unstable periodic orbits which shift between the relative spheres of influence of different bodies.
Note that the above discussion really focuses on the energy state rather than speed or velocity. This is because in celestial mechanics, the position and velocity state vectors with respect to an epoch is a calculated result that is of interest for specific mission parameters (such as the position of a satellite relative to a ground station or trajectory of spacecraft during a flyby), but all of the orbital simluations are done in terms of energy and the the Keplerian orbital elements (or for non-periodic trajectories, the universal form) as these are in the natural forms for orbital mechanics. Direct iterative simulation of state vectors is only done in the case of applied impulse (such as spacecraft thrusting or momentum change from impact) or for the case that there are fine perturbances which are best represented by change in velocity rather than a change in energy state.
Stranger
you return to the point of “firing” if you don’t have escape velocity. If this is straight up, thn the object comes straight down again (elliptical orbit with basically zero minor axis).
Of course, you have to take the velocity of earth’s rotation at launch point into consideration. You aren’t shooting straight “up” if the gun is also rotating at almost 1000mph, you’re shooting at a slightly sideways angle. If it’s a rocket, and the velocity grows with time until engine cutoff, the math is a bit more complex, but essentially - it’s whatever ellipse is defined by that velocity and direction.
The simplistic way of looking at it - if you accelerate tangent to the force of gravity - that defines the perigeegee (low point) of the ellipse. Given the right velocity and no air resistance, you can fire a cannon at circular orbital velocity and the cannonball would circle the globe and hit the back of the cannon (again, minus earth rotation effects.)
Fire less than circular velocity, and the cannon defines the apogee, or high point in the ellipse, the cannon ball will fall through the earth and come back to the canoon. Minus, again, friction effect of the cannonball going through the earth, rotation of the earth, and the pesky detail that gravity decreases as you penetrate the earth until at the center of the earth you would be theoretically weightless…
If you fire a cannon horizontally but greater than circular orbit speed, the cannon is sitting at the low spot (perigee) of the orbit. minus… yada yada…
AKA “lithobraking”.
Of course when you actually launch with a plan to leave the planet (or moon, or star) behind you realise that if the rocket fails , you are not so much in orbit , but really will just fall back to where you came from from…
That is, the launch to escape is likely to be totally unlike a launch to a nice orbit.
I wanted to elaborate a bit, and finally got around to it. If you start in a circular orbit just skimming the earth’s surface, the orbital velocity is about 5 miles a second, orbital radius is 4000 miles, and orbital period is 90 minutes (all numbers kept to 1 significant digit, just for convenience). Escape velocity is 7 miles a second.
If you boost the orbit by suddenly increasing the tangential velocity from Vo to V, you will create a new orbit with its lowest point as the current orbital radius, and a new highest altitude of 4000 miles times Vo^2/(2Vo^2-V^2), and the new period is 90 minutes times (Vo^3)/((2Vo^2-V^2)^(3/2)), so
For V=1.05Vo, the highest altitude is 1.14000 miles and the period is 1.2 90 minutes
For V=1.1 Vo, the highest altitude is 1.34000 miles and the period is 1.4 90 minutes
For V=1.15Vo, the highest altitude is 1.54000 miles and the period is 1.8 90 minutes
For V=1.2Vo, the highest altitude is 1.84000 miles and the period is 2.4 90 minutes
For V=1.25Vo, the highest altitude is 2.34000 miles and the period is 3.5 90 minutes
For V=1.3Vo, the highest altitude is 3.24000 miles and the period is 5.8 90 minutes
For V=1.35Vo, the highest altitude is 5.64000 miles and the period is 13.4 90 minutes
For V=1.4Vo, the highest altitude is 254000 miles and the period is 125 90 minutes
The closer you get to escape velocity, the highest altitude and the period both grow faster and faster towards infinity.
