How to explain the trajectory of object going up then come down with curvature of spacetime? I can understand that the Moon is orbiting the Earth because it is going in a straight path within the distortion of spacetime caused by the Earth's mass, so not outside force is required. However, if I throw an object straight up, then how does the curvature of spacetime make it come down without influence of outside force? I know I'm applying a force to the object upwards, so it should stay in geodesic motion, but why does it fall?
 A: In both situations the objects (the Moon or the object you threw up) are moving in geodesic motion.
In the framework of General Relativity, gravity is not a force, but rather an expression of the curvature of spacetime. When an object is in free fall (for example, after you have let it go from your hand when throwing it up, if we ignore air resistance), it is subject only to the effects of the curvature of spacetime. Just like the Moon, it follows the straightest possible path on spacetime, which is geodesic motion. However, when we see something fall, we usually do not think about its trajectory on spacetime, only about its trajectory on space as a function of time. When one considers the geodesic motion on spacetime and uses it to compute the trajectory in space as a function of time, one will get the results we observe everyday (i.e., roughly a parabola), plus some corrections due to relativistic effects and/or the fact that the Earth is round.
Hence, the object falling is in geodesic motion and, in the general relativistic framework, it is not under the effects of any forces whatsoever (as long as we ignore air resistance and etc). Of course, we clearly see the effects of gravity, but gravity is not a force in General Relativity, and what we see are merely the consequences of the fact that the geodesics in a curved spacetime are highly non-trivial.
