Thrust Force on Rocket

Derivation

From the rocket equation derivation:

$M \, dv = -u \, dM$

Divide by $dt$:

$M\frac{dv}{dt} = -u\frac{dM}{dt}$

$Ma = -u\frac{dM}{dt}$

The left side is $Ma$ — force by Newton's Second Law. The right side is the thrust:

$\boxed{F_{thrust} = -u\frac{dm}{dt}}$

Since $\frac{dm}{dt} < 0$ (mass is decreasing), $F_{thrust} > 0$ — thrust is positive (forward).

Physical meaning

The thrust force is the reaction to the momentum carried away by the exhaust. By Newton's Third Law:

• Rocket pushes exhaust backward at speed $u$ relative to rocket
• Exhaust pushes rocket forward with equal and opposite force

Rate of momentum carried away by exhaust per unit time $= u \times |\frac{dm}{dt}|$ = thrust.

Net force on rocket

In gravity, with air resistance $f$:

$F_{net} = F_{thrust} - mg - f = -u\frac{dm}{dt} - mg - f = Ma$

$a = \frac{-u\frac{dm}{dt} - mg - f}{m}$

As fuel burns and $m$ decreases, acceleration increases — even if thrust is constant.

Example

A rocket ejects exhaust at $u = 2000$ m/s at a rate of $\frac{dm}{dt} = -10$ kg/s:

$F_{thrust} = -2000 \times (-10) = 20000 \text{ N} = 20 \text{ kN}$

If the rocket currently has mass $m = 500$ kg (no gravity):

$a = \frac{F_{thrust}}{m} = \frac{20000}{500} = 40 \text{ m/s}^2$

Condition for liftoff

For a rocket to lift off vertically, thrust must exceed weight:

$F_{thrust} > mg$

$-u\frac{dm}{dt} > mg$

$\left|\frac{dm}{dt}\right| > \frac{mg}{u}$

The rate of fuel consumption must be large enough that thrust exceeds gravity.

Multi-stage rockets

The thrust formula shows that $F_{thrust} = u|\dot{m}|$ is determined by exhaust speed and burn rate — independent of the rocket's current mass.

As fuel burns and $m$ decreases, $a = F_{thrust}/m$ increases. This is why rockets accelerate faster toward the end of a burn.

Discarding empty stages removes dead mass, improving the acceleration in subsequent stages. This is the key advantage of multi-stage rockets over a single-stage design.