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What came to my mind is to throw an object far from the earth with a particular speed equal to $$V_{sat}=\sqrt{\frac{G\times M_{central}}{R}}$$like satellites. However, it would follow a full circle.

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  • $\begingroup$ You have to specify the height at which it is launched in order to find out when it is going to hit the ground. $\endgroup$ Nov 9, 2013 at 16:53

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In general, a particle under a central force will follow a conic. The only possibility of it following a full circle is to have the perfect direction and speed so that the falling down is compensated by the going forward. Indeed, like a geostationary satellite.

However, that motion would be perpetual in absence of friction. For it to follow a half circle only, it has to be stopped by something.

Remember though that Earth, for example, is not completely spherical. For simplicity, imagine it is a spheroid, which is close to reality. The Equatorial radius is slightly longer than the polar radius.

If you launch the object near the Equator, it can follow a circle with a radius slightly larger than the polar radius, but slightly smaller than the equatorial. Therefore, it will go over the Pole but will crash at the other side of the Earth, at the same distance from the Equator that it was launched. The net effect, if done under the right conditions, is that it will have described a half circle.

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  • $\begingroup$ Certainly. However that is a deviation from the simplification that I was going for here, which is the spheroidal Earth. In any case, you are right. $\endgroup$
    – legrojan
    Nov 9, 2013 at 21:39
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This is 1st order approximation, but find the time $t$ to drop a certain height $h$ under approximately constant gravity, and the find the speed needed to go around the 1/2 earth in this time.

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