Gravity Drag - Example

Example

Consider the simplified case of a vehicle with constant mass accelerating vertically upwards with a constant thrust per unit mass a in a gravitational field of strength g. The actual acceleration of the craft is a-g and it is using delta-v at a rate of a per unit time.

Over a time t the change in speed of the spacecraft is (a-g)t, whereas the delta-v expended is at. The gravity drag is the difference between these figures, which is gt. As a proportion of delta-v, the gravity drag is g/a.

A very large thrust over a very short time will achieve a desired speed increase with little gravity drag. On the other hand, if a is only slightly greater than g, the gravity drag is a large proportion of delta-v. Gravity drag can be described as the extra delta-v needed because of not being able to spend all the needed delta-v instantaneously.

This effect can be explained in two equivalent ways:

• The specific energy gained per unit delta-v is equal to the speed, so spend the delta-v when the rocket is going fast; in the case of being decelerated by gravity this means as soon as possible.
• It is wasteful to lift fuel unnecessarily: use it right away, and then the rocket does not have to lift it.

These effects apply whenever climbing to an orbit with higher specific orbital energy, such as during launch to Low Earth orbit (LEO) or from LEO to an escape orbit. This is a worst case calculation - in practice, gravity drag during launch and ascent is less than the maximum value of gt because the launch trajectory does not remain vertical and the vehicle's mass is not constant, due to consumption of propellant and staging.