WHY is the energy required to launch a rocket vertically to a height h , less than the energy?

Question

WHY is the energy required to launch a rocket vertically to a height h , less than the energy required to put the same rocket into orbit at the height h?

Answer 1

To put it in orbit requires both vertical (height h) and horizontal motion. Horizontal motion would require additional energy.

Answer 2

Both require mgh = PE energy to get to the same height. What is different is the power required to lift the rockets by the two different trajectory profiles. The straight up trajectory will use more power than the gradual trajectory.

Going up gradually to slip into an orbit is like going up a mountain side using a switchback road that winds back and forth. Going straight up is like climbing that same mountain, but on its shear cliff side. In both cases, the potential energy at the top of that mountain will be PE = mgh; where m is the mass, g is acceleration due to gravity force, and h is the mountain peak height.

That the energy (work) it takes to get there at h is the same can be proved through a thought experiment. Suppose two people climb the mountain to its peak at h. One person goes the windy road, the other climbs the cliff. Now both of them drop a golf ball over the cliff at the same time. Which ball will get to the ground first? Answer...neither, they'll both arrive back on the ground at the same time. They will both have a kinetic energy on impact equal to PE = mgh = 1/2 mv^2 = KE So the amount of energy it took to carry the mass up the windy road was the same as it took to climb the cliff.

Power is different though. Power P = E/det t; where E is energy and del t is the time elapsed expending that energy. As you might guess, the going straight up trajectory will use more power than the gradual trajectory using the same energy. This results because del t for the straight up rocket will be less than the del T for the gradual trajectory. The gradual del T is greater to get to h because the vertical velocity is less than the vertical velocity of the straight up trajectory. Let the vertical velocity be v = V sin(theta); where theta is the angle of incline for the gradual trajectory and V is the velocity of the rocket. Theta = 90 deg for the straight up rocket and theta < 90 deg for the gradual trajector. Thus, v = V for the straight up and v = V sin(theta < 90) < V for the gradual.

Since h = v del t; so that del t = h/v, we see that del t = h/V < del T = h/v when V > v. Therefore, del t < del T; so that P = E/del t > E/del T = p which is the lesser power used to raise the rocket to h on an inclined trajectory.

Lesson learned, it takes the same amount of work (energy) to raise an object to a height h by a gradual route as it does to raise that object straight up to h. On the other hand, it takes more power to raise the rocket straight up than it does to climb gradually. (On a practical note, that power differential is a major reason you don't see aircraft take off straight up.)

Answer 3

Both require the same potential energy - the height is the same.

However, if the rocket goes straight up then when it reaches height h, it just falls back. To stay in orbit, it needs enough kinetic energy to balance the potential energy it has at height h - or GMm/(R + h) = 1/2 mv^2.