Mathematical Formulae for Astronomy
Newton's Formula for Orbital Velocity involves knowing the mass of the object
you wish to orbit and the distance from the center of that object. This formula
grew out of Newton's curiosity about the orbit of the moon. How could the moon
travel in a curved path around the Earth and neither crash downupon us nor fly
away from us? To Newton, the moon was falling toward the Earth at the exact
same speed that it was moving forward in a straight line. This caused the path
of the moon to be circular, and with no other force acting upon the moon, it
When William Tell shot the apple, his arrow did NOT travel in a straight line,
but curved downward due to the gravitational attraction of the Earth upon the
flight of the arrow. A great major league pitcher puts spin on the baseball
to effect a curveball, when in actuality, the ball is curving more downward
under the force of gravity upon the speeding ball, and the poor batter swings
and misses. An expert marksman must take the effect of gravity into account
when firing a bullet at a target several hundred meters away by aiming above
the target and knowing the velocity of his brand of bullets.
When Sputnik flew overhead in October, 1957, the race was on to put satellites
into orbit. Since that initial moment of panic, engineers from numerous companies
try to put things in orbit for various reasons. Let's look at the speed that
the NASA Shuttle must go in order to orbit at an altitude of 200 km, which is
a common orbital altitude.
Newton's formula for Orbital Velocity is
Vc is the orbital velocity
m is the mass of the object you wish to orbit ... measured in kilograms
r is the radius of the orbit around the center of the object ... measured
g is Newton's Constant of
The radius of Earth is ,
and the mass of Earth is .
Since the shuttle orbits at 200 km, we need to add 200,000 meters to the radius
of the Earth.
= 7786.94 m/s = 7.787 km/s, which is 17,380 miles/hr. While this speed is not
enough to reach escape velocity, it is sufficient
to put the shuttle in orbit.
So, let's imagine that we wish to determine the circular velocity of a satellite
that we wish to launch to an altitude of 36,000 km. To determine what this speed
is, we need to know a few values. First, the mass of the Earth is ,
and the radius of Earth is .
Since we are launching the satellite to an altitude of 36,000 km, we are actually
launching it to a point 36,000 km + 6378 km above the CENTER of the Earth. This
means, we are launching it 42,378 km from the center of the Earth. Once again,
we need to convert all values in Newton's equations from kilometers into meters,
so the formula looks liike:
= 3067.89 m/s = 3.06789 km/s.
This number looks pretty irrelevant, and yet it is VERY important. The importance
is discovered by looking at a few more numbers. The total circumference of the
Earth at 36,000 km is 2r,
where r =
= 266,268,602 m or 266,268 km. A satellite that travels at 3.0678 km/sec will
cover this distance in distance is just about 24 hours! This means, the satellite
will NOT appear to move in the sky during the entire night, but APPEAR to be
fixxed in space. It is this interesting orbital location that has made satellite
tv and cellular phone service so much more simple. In my childhood, radars were
seen as antennae that spun in circles. Now your satellite dish affixed to your
rooftop is pointed at a fixed spot in space (the location of the satellite transmitter)
and never needs to be adjusted again ... un less a big wind blows it out of
alignment. The satellite has been launched to GEOSTATIONARY orbit.
What is important to recognize here is the importance of radius relative to
mass. Keeping mass the same, but reducing or increasing the altitude has a dramatic
effect on the orbital velocity. Why not try to figure out the orbital velocity
of a golf ball on Mars' moon Phobos.
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