Educator's Guide to The Inverse Square Law
Courtesy of the Jet Propulsion Laboratory
How Far/How Faint
How much brighter is the Sun as viewed
from the planet Mercury as
compared to Earth? How much fainter
is it at Neptune? How
strong is the Sun's gravitational pull on the
spacecraft now as compared to when it was at
Jupiter? How much
pull does the Sun exert on the nearest star?
Actually, all of these questions can be answered through a very
simple mathematical relationship known as the inverse square law.
It is a simple division problem that can be applied to a variety
of interesting situations that affect planets, spacecraft that
venture out into deep space and a number of other natural
The equation relates the relative distances of two objects as
compared to a third. Typically one of the objects is Earth, the
second is a spacecraft and the third is the Sun. To begin, lets
make some generalizations. There is a certain amount of sunlight
reaching Earth at any given moment. This is not an absolute
quantity because Earth is closer to the Sun at some times of the
year verses others and the number of sunspots effects the Sun's
Overall, however, the Sun is remarkably constant in its
behavior. If it were not, life on Earth might be impossible.
We can describe the amount of the Sun's energy reaching Earth as
1 solar constant. The average distance from the Sun to Earth is
149,597,870.66 kilometers (92,955,807.25 miles),
which we can
simplify to what astronomers call 1 Astronomical Unit or 1 AU.
So Earth is 1 AU from the Sun and receives 1 solar constant.
This will help keep the math easy.
The relationship can be expressed most simply as: 1/d^2 (one over
the distance squared) where d = distance as compared to Earth's
distance from the Sun (for our first examples).
Let's start with sunlight as an example. At 1 AU, Earth receives
1 unit of sunlight; what we generally might associate with a
bright sunny day at noon. How much sunlight would a spacecraft
receive if it were twice as far from the Sun as Earth? Your
first guess might be that, since it is twice as far it will only
receive half as much (not twice as much since it is farther
The distance from the Sun to the spacecraft would be 2 AUs so...
d = 2. If we plug that into the equation 1/d^2 = 1/2^2 = 1/4
= 25% The spacecraft is getting only one quarter of the amount of
sunlight that would reach it if it were near Earth. This is
because the light is being radiated from the Sun in a sphere. As
the distance from the Sun increases the surface area of the
sphere grows by the square of the distance. That means that
there is only 1/d^2 energy falling on any similar area on the
Now lets try it for another real place. Mars is at a distance of
1.5 AUs from the Sun. 1/d^2 = 1/1.5^2 = 1/2.25 = 44%. There
is less than half as much sunlight falling on the surface of Mars
as on Earth! Jupiter is at 5.2 AUs so 1/d^2 = 1/5.2^2 = 1/27
= 3.7%. Neptune is at 30 AUs so 1/d^2 = 1/30^2 = 1/900 Å
0.1%! Noon on Neptune is like very deep twilight on Earth!
What happens as we approach the Sun? Common sense tells us that
the Sun will be brighter and the inverse square law tells us how
much brighter. Mercury is at 0.387 AUs. 1/d^2 = 1/0.387^2 =
1/.15 = 666.67%, almost seven times brighter! We can use this
method to compare any spot in the Universe if we describe its
distance as compared to Earth relative to the Sun.
Planet Dist. (in AUs)
Pluto (min.)* 29.69
Pluto (avg.) 39.44
Pluto (max.) 49.19
a Cen** 272,000
* Pluto's eccentric orbit carries it closer to the Sun than
Neptune, where it is now and will be until March 1999.
** Alpha Centauri - the star system (three stars) nearest our
Sun; approximately 4.3 light years away (63,240 AU/light year).
Not visible from
the United States except in Hawaii.
We have been comparing sunlight but this is exactly the same
method that we would use for any other form of randomly radiated
energy such as heat, ultraviolet or x-rays, magnetic field
strength or gravity. The gravitational tug that the Sun exerts
on Earth can be compared to the Sun's tug on Mercury, Pluto,
spacecraft or the stars.
Continue on to learn how to make more calculations.
-- "Zero Gravity?" Not in this universe!
Everyone of us is familiar with gravity. It is as simple as
falling down. But whenever we start to discuss space the concept
of gravity becomes relative and confusing. This is especially
true on manned space flights where the true gravitational status
of Space Shuttle astronauts or cosmonauts onboard the Mir space
station is often described incompletely.
What terms are correct in
which contexts? The most important term
of all is gravity. Gravity, or gravitation, is the attractive
force between bodies related to their masses and distances.
Every object with mass exerts a gravitational pull on every other
massive object. The more massive the object, the stronger the
gravitational attraction. The closer the objects, the more
strongly the attraction will be felt.
Gravity was first quantified by Sir Isaac Newton (he did not
"discover" it). He realized that the same force that caused
apples to fall from trees to Earth (there is a popular legend
associated with this revelation) was responsible for holding the
Moon in orbit around Earth and the planets around the Sun. He
also realized that as distance grows the felt attraction drops
but never reaches zero. In other words, every massive object in
the Universe exerts a gravitational attraction on every other
massive object in the Universe!
So how does the gravity of Earth affect a spacecraft in orbit?
In fact, the gravitational pull of Earth on the shuttle and the
astronauts onboard is almost exactly the same as the
gravitational pull holding you in your seat right now. The
astronauts are not in "Zero G" ("G" or "Gee" is an abbreviation
for gravity). We will prove this shortly.
The astronauts are weightless and they are in free fall. Newton
realized that gravity's effects on objects could be described in
terms of falling. Apples fall from trees and the Moon falls
around Earth. If you stand out on a field and throw a baseball
directly towards the horizon it would travel indefinitely if it
were not being acted upon by outside forces. One of these forces
is friction produced by the resistance of the air that the ball
is passing through. Another is the resistance of the fence or
trees at the edge of the field.
But even in the presence of these forces, the most conspicuous
force acting upon the ball is gravity. It causes the ball to
fall to the surface in an arc. But what if you throw the ball
harder? The ball will travel farther but will still fall in an
If we negate the air resistance, and the fence, and really give
the ball a heave we will be able to throw the ball so far that as
it arcs towards the surface,
the surface arcs out of its way.
After all, the surface of Earth is curved. If we use a canon, or
rocket motors, we can get that ball going so fast that the arc of
its fall exactly matches the arc of Earth's surface! This way
the ball is continually arcing towards a surface that is
continually arcing out of its way.
The speed necessary to reach this situation is about 28,200
kilometers per hour (17,500 miles per hour) and this is the speed
that the Space Shuttle must attain to continue to fall around
Earth (remain in orbit) at an altitude of 300 kilometers (186
miles). Therefore the shuttle and the astronauts onboard are in
free fall but still in the gravitational pull of Earth. If
Earth's gravity were to somehow disappear, they would fly off in
a straight line at a tangent to their orbit.
How much weaker is the force of gravity at 300 kilometers above
Earth's surface? We will use an equation with four
multiplications and one division:
- R =
- the radius of a massive object;
- r =
- the distance from the center of mass of the massive object to
the center of mass of a much smaller massive object;
- g =
- the gravitational attraction of the massive object on objects
on its surface;
- and gr =
- the gravitational attraction of the massive object as
felt on the less massive object.
For Earth, R = 6378.2 km.; r = 6378.2 + 300 km.; g = 1 gee. Plug
in the numbers to see how many gees the astronauts are actually
experiencing. If they did not know better they might think that
they were in 0g but we know better (problem answer below). (The
equation can be modified for other planets by replacing "R" with
the new planet's radius and "g" with its surface gravity. For
example, for Jupiter: R = 71,398 km. and g = 2.64g.)
Answer to above problem:
gr = 0.912 G or 91.2% gravity felt at Earth's surface.