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Assuming the earth had no atmosphere, and you could throw a baseball as fast as you wanted to, and you were standing on the surface of the earth, is it possible to throw the baseball so fast that it circles the earth and flies over your head above the height you threw it at?

The thinking goes like this: you throw a baseball and it lands far away from you. Then you throw it harder and it lands half way around the world. Then you throw it harder and it lands at your feet from behind. Then you throw it harder and it passes your head, in orbit.

But could you throw it even faster so that it passes you above your head?

I think this is not possible, because if you threw it even harder, the orbit would just become elliptical and the ball would still arrive at your head height. Even if you angled it slightly up or down, it would still only arrive back to your at head height. Right?

It's because baseballs are not rockets. They can't circularize their orbits once they have some altitude. Right?

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    $\begingroup$ So the solution is to throw like volleyball high style instead of baseball style? Is it so hard to throw a ball at above head height? $\endgroup$ – user3528438 Feb 1 at 15:24
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    $\begingroup$ How simplified do you want the problem to be? the Earth does not have uniform gravitational strength, leading to minor wobbles in orbit (in all 3 dimensions). $\endgroup$ – Carl Witthoft Feb 1 at 18:25
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    $\begingroup$ Are you assuming a spherical Earth? $\endgroup$ – Mark Feb 2 at 0:01
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    $\begingroup$ If you could throw it as fast as you wanted too, you could aim it to bounce of some debris to circularise. $\endgroup$ – user20636 Feb 2 at 0:10
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    $\begingroup$ You can release the ball above the height of your head. Most people have arms that they can raise above their head. $\endgroup$ – Polygnome Feb 3 at 12:11
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Yes, but only because the Earth rotates. If you throw the ball, you will end up in a slightly different spot. If you were at a high point at the equator and threw the ball due East at higher than orbital velocity, you would rotate by the time the ball came back and it would be slightly over your head, because you have moved to a different spot in the orbit. The lowest point in the orbit, however, would be where you released the ball.

If Earth also didn't rotate, it would return at whatever point you released the ball at. Most people don't throw a ball over their heads, so...

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    $\begingroup$ So if you tried to repeat your performance from the same time and place on the next day, you might get clobbered by the first ball. But any other time of day you should be safe. $\endgroup$ – A. I. Breveleri Feb 2 at 2:24
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    $\begingroup$ But if you are anywhere but on the equator, or throw the ball in any direction but due east (or west), the ball will not return to where you're standing, because the Earth will have rotated about 22.5 degrees during the orbit. (Assuming one orbit takes 90 minutes.) $\endgroup$ – jamesqf Feb 2 at 2:40
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    $\begingroup$ This answer is assuming that, in addition to lacking an atmosphere, the Earth is perfectly round and homogeneous (except for radial density variation). Bumps on the actual, realistic, Earth would alter the trajectory, making it non-closed and thus changing the "same-point" outcome. $\endgroup$ – Ruslan Feb 2 at 10:55
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    $\begingroup$ Upvoted for "Most people don't throw a ball over their heads" - I have yet to find a more convincing proof of any theorem $\endgroup$ – Mawg says reinstate Monica Feb 2 at 13:50
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    $\begingroup$ Nitpick: the lowest point on the orbit would only be where you released the ball if you threw it perfectly horizontally. If it was angled slightly up or down at the release point (as the OP suggests), then its perigee would be closer to the center of the Earth than the release point was. $\endgroup$ – Michael Seifert Feb 3 at 14:58
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You're correct.

On a perfectly spherical, atmosphere-free Earth, with no obstacles as tall as you, with a uniformly spherical gravitational field, it would be possible; the low point of the orbit would be at the altitude you threw the ball from, a couple of meters above the surface.

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    $\begingroup$ Provided it was thrown horizontal, and that's not guaranteed. When I throw a baseball the trajectory is pretty random, it's basically a Monte-Carlo method for estimating the cross-sectional area of the batter. $\endgroup$ – uhoh Feb 1 at 19:05
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    $\begingroup$ If the ball could go through the earth without friction, then even if you throw the ball up or down, it'd still orbit and end up right at the point where you released it $\endgroup$ – Mooing Duck Feb 1 at 23:29
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    $\begingroup$ @uhoh I figure if we're granting the ability to throw a ball at 7800m/s, we can throw in "good aim" for free. $\endgroup$ – Russell Borogove Feb 1 at 23:33
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    $\begingroup$ The OP has independently thought of Newton's Cannonball, If a ball is thrown horizontally hard enough from the top of a mountain it will hit the back of the cannon after completing one orbit $\endgroup$ – CSM Feb 2 at 10:50
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    $\begingroup$ @uhoh When my partner is throwing a ball for the dog, a 100% success rate in the ball even going forwards would be a noticeable improvement, at least from the point of view of those who want to avoid getting hit by a tennis ball covered in saliva. $\endgroup$ – Graham Feb 2 at 16:49
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"...above the height you threw it at?"

No. It would come back to exactly the height* from which you threw it, unless you threw it so hard (i.e., at "escape velocity") that it did not come back at all.

All orbits are conic sections (ellipses, parabolas, hyperbolas.) If you throw the baseball, at exactly escape velocity, then it will follow a parabolic trajectory that never returns. If you throw it any harder, it will follow a hyperbolic trajectory, and if you throw it less hard, it will follow an elliptical trajectory.

The elliptical trajectories all return to their starting point, but the shape of the ellipse (i.e., it's "eccentricity") and how long it takes to return depends on how hard you throw it.


* Not just exactly the same height, but also, if you're plotting its position in an inertial coordinate system that is at rest with respect to the common barycenter of the baseball and the planet, and if no other forces besides gravity act on it, then it will come back to exactly the same place where it left your hand.

But also note, if the planet is rotating, then it's highly likely that you will no longer be in that place when the ball comes back to it.

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you can't throw it so that it lands behind your feet. if it doesn't hit halfway around the world, it will return to your hand, because elipse

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A circle is a subset of ellipse, and a circular orbit can (on a theoretical, magical, non-rotating, vacuum Earth) be thrown. A slightly slower orbit might reasonably be thought to hit your knees (to avoid hitting the antipodal soil), but it would no longer be circular but again elliptical, with a greatly shifted major axis, since you would be releasing it at its apogee rather than perigee.
You might also note that you can slightly shift the elliptical axes of this almost circular orbit, by aiming slightly above or below your shoulder at the 'horizon' (minor axis).

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  • $\begingroup$ I think a slightly [s]lower orbit (thrown horizontally, but slower than needed for a circular orbit) would still result in an elliptic trajectory, but now you'll be on top of the major axis, and the ball will hit the ground 90 or fewer degrees in front of you because the minor axis is shorter than Earth's radius. Actually, every ball you throw follows an elliptic trajectory (minus air resistance) which would take it close to the Earth's center and back, if it could (and Earth's mass were concentrated there). $\endgroup$ – Peter - Reinstate Monica Feb 3 at 12:13
  • $\begingroup$ oh, you're right. clearly seen. $\endgroup$ – amara Feb 3 at 16:21
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In addition to the argument about the earth rotating, it is also possible because you don't release the ball from your head, you release it from your hand. By the rules of 2-body problem orbits, the orbit will always intersect with the release point. This point is in front of you. So in theory one could have an elliptical orbit which is at a higher altitude above your head before intersecting with your hand.

This means you need to throw the ball downward, not horizontal or up. That will cause the ball to be coming down when it completes the orbit. This of course, is a challenge because the earth is going to get in the way of your ball. If it wasn't there, it would be easy to achieve your goal.

As such, I would say you need to climb Mt. Everest, and throw the ball with a slight downward angle. You also should release the ball as high as you possibly can. I don't think this can be done if you release at shoulder height, just due to geometry. To have the path go above your head and then hit shoulder height at the release point would require a very steep angle... meaning you had to throw it downrward more -- and you're more likely to intersect the earth.

Beyond that, it's just a geometry game. You need a pitch that's high enough (near head level) that when you look at the perigee of the orbit, it doesn't intersect the ground.

On the other hand, if we start asking how you manage to accelerate the ball that fast, we might be talking about a long robotic arm. If its long enough, that could make the geometry easier because it permits a shallower angle while still being able to go overhead.

Of course, the easy answer is to throw the ball goofy, holding you hand up above your head and releasing it there.

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As long as the ball is thrown at escape velocity (~11 km/s) its curve due to falling under gravity will be the same as the circumference of the earth - this is what a circular orbit is.

Of course this assumes a perfectly spherical earth of constant density and no atmosphere (none of which are true), but if these were true and one could throw at such crazy high speed in an exactly tangential direction, then the ball will continue at constant height above the perfectly spherical earth, eventually returning to the same position and height.

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    $\begingroup$ Your velocity number is incorrect. 11 km/s will result in a parabolic (or hyperbolic) trajectory, not a circular trajectory. If an object is above escape velocity, it will never circle the Earth, it will move away for infinity. The orbital velocity is roughly 7.6 km/s. $\endgroup$ – Star Man Feb 26 at 2:08
  • $\begingroup$ @StarMan you are incorrect and thus your down voting is not appreciated. From en.wikipedia.org/wiki/Escape_velocity: The escape velocity from Earth's surface is about 11,186 m/s (6.951 mi/s; 40,270 km/h; 36,700 ft/s; 25,020 mph; 21,744 kn). $\endgroup$ – Socco Feb 26 at 3:13
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    $\begingroup$ Escape velocity is the speed needed to escape (the earth) and never return, thus it will never orbit, because it will never return. $\endgroup$ – Magmatic Feb 26 at 3:25
  • $\begingroup$ Ok, I concur. You are correct, and I have now learnt the difference between escape velocity and orbital velocity. $\endgroup$ – Socco Feb 26 at 3:34
  • $\begingroup$ @Socco a) I didn't downvote. b) escape velocity != orbital velocity $\endgroup$ – Star Man Feb 26 at 16:12

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