Knowing Planet Distance from Light Visibility

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 Voyager 1 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 phenomena.

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 since Earth is closer to the Sun at some times of the year verses others and the number of sunspots effects the Sun’s energy output, but overall the Sun is remarkably constant in its behavior. If it were not, life on Earth might be impossible.

Image above: The Intensity (or brightness) of a source of light falls off as the inverse square of the distance from the source. This is because the light’s intensity can be thought of as being spread out uniformly over the surface of a sphere. At greater distances, the sphere is larger, so the same amount of light is spread over a larger area. This makes the light source appear dimmer as the distance from it increases. Credit: NASA

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 away).

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 expanding sphere.

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 ”° 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.

* 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 U.S. 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.

Gravity Vocabulary… “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 what 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!

Image to right: Scientists use mathematics to determine how gravity will affect astronauts and spacecraft. Credit: NASA

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 arc.

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:

gr = R^2/r^2 x g

where: 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.)

gr = R^2/r^2 x g

Answer to above problem:

gr = 0.912 G or 91.2% gravity felt at Earth’s surface.