The forces on a rocket change
dramatically during a typical
This figure shows the forces and resulting velocity of a rocket
while descending through the
The velocity at any time during the flight depends on
the corresponding acceleration of the vehicle and the
balance of forces acting on the vehicle.
Forces, accelerations, and velocities are all
having both a magnitude and a direction. When describing the
motion of an object caused by forces, one must
both the magnitude and the direction.
During coasting flight,
are produced in response to Newton's
The forces present during coasting flight are the
and the weight.
The weight is constant in magnitude and is always
directed toward the center of the earth.
The magnitude of the drag changes with the
square of the velocity.
The direction of the drag
is along the rocket axis and opposed to the motion
of the rocket. During
the drag is in the same direction as the weight.
But during descent, the drag opposes the weight.
For a descending rocket,
the net vertical external force F is equal to the
difference between the
drag D and the weight W:
Drag increases with the
square of the speed.
So as the rocket falls, we quickly reach conditions where the
drag becomes equal to the weight, if the weight is small.
When drag is equal to weight, there is no
net external force on the object
and the vertical acceleration goes to zero. With no acceleration,
the object falls at a
constant velocity as described by Newton's
of motion. The constant vertical velocity is called the terminal
Using algebra, we can determine the value of the terminal velocity.
At terminal velocity:
D = W
Cd * r * V ^2 * A / 2 = W
Solving for the vertical velocity V, we obtain the equation
V = sqrt ( (2 * W) / (Cd * r * A)
Typical values of the drag coefficient are given
on a separate slide.
For a model rocket, the value is near 0.75.
Because model rockets do not fly very high, the air density is nearly
constant and equal to the sea level value
(.00237 slug/cu ft or 1.229 kg/cu m) on the surface of the
For a rocket landing on
Mars, the atmospheric density is quite low,
and the terminal velocity correspondingly higher. But the
weight on Mars is also lower than on Earth
so the terminal velocity is not as high as predicted by the atmospheric
Here's a Java calculator which will solve the
equations presented on this page:
Due to IT
security concerns, many users are currently experiencing problems running NASA Glenn
educational applets. There are
security settings that you can adjust that may correct
As mentioned above, the atmosphere and gravitational constant of a planet
affects the terminal velocity. You select the planet using the choice button
at the top left. You can perform the calculations in English (Imperial) or
metric units. You must specify the weight or mass of your object. You can
choose to input either the weight on Earth, or the local weight, or the mass
of the object. Then you must specify the cross sectional area and the drag
coefficient. Finally you must specify the atmospheric density. We have included
models of the atmospheric density variation with altitude in the calculator.
When you have the proper test conditions, press the red "Compute" button to
calculate the terminal velocity.
You can download your own copy of this calculator for use off line. The program
is provided as TermVel.zip. You must save this file on your hard drive
and "Extract" the necessary files from TermVel.zip. Click on "Termvcalc.html"
to launch your browser and load the program.
Notice In this calculator, you have to specify the
The value of the drag coefficient depends on the
of the object and on
in the flow.
For airflow near and faster than the
speed of sound,
there is a large increase in the drag coefficient because of
the formation of
on the object. So be very careful when interpreting results with large
terminal velocities. If your drag coefficient includes compressibility
effects, then your answer is correct. If your drag coefficient
was determined at low speeds, and the terminal velocity is very high,
you are getting the wrong answer because your drag coefficient does
not include compressibility effects.
The terminal velocity equation tells us that an object with a
small cross-sectional area, or a low drag coefficient, or a
heavy weight will fall
faster than an object with a large area, or high drag coefficient,
or a light weight. A
rocket with a small parachute will fall faster than with a
large parachute because of these effects.
Since a rocket will drift with the wind, you can make your
rocket return closer to the pad on a windy day by using a
If we have two objects with the same area and drag
coefficient, like two identically sized spheres, the heavier object
falls faster. This seems to contradict the findings of Galileo that
all free falling objects fall at the
same rate. But Galileo's principle only
applies in a vacuum, where there is NO drag.
Any two objects fall at the same rate on the
Moon because there is a vacuum and no drag
and only gravity acting on the objects. In general, the same two objects fall at
different speeds on the Earth and on Mars because of aerodynamic drag.