To escape Earth's gravity from rest, ignoring air resistance and gravity's effect during the burn, a rocket needs a mass ratio m0 over mr such that the natural logarithm of that ratio equals 4.48. What percentage of the rocket's initial mass must be fuel in this idealized case?

1. Solving ln(m0 over mr) equal to 4.48 for the ratio itself gives m0 over mr equal to e raised to 4.48, which is about 88. 2. That means only 1 out of every 88 parts of the initial mass remains once the fuel is burned, so mr equals m0 divided by 88. 3. The fuel mass is m0 minus mr, which is m0 times (1 minus 1/88). 4. Dividing by m0 and converting to a percentage gives about 98.9 percent of the rocket's initial mass as fuel. 5. 88 is the mass ratio itself, not the fuel percentage, and 1.1 percent is instead the small leftover share taken up by payload, engines, and tanks. _Source: OpenStax College Physics (CC BY 4.0), Ch 8 "Linear Momentum and Collisions", section 8.7 Introduction to Rocket Propulsion_