If you asked random people why things in orbit don't fall I'd expect you'd get an answer about lack of gravity in space. While strictly speaking there is so little gravity in the majority of space as to be considered none, this is not true of near Earth orbit. Things like the space shuttle and the International Space Station orbit at about 200 miles above the surface of the Earth.

Orbital mechanics can be pretty complex. There are strange results like it taking more energy to send a probe closer to the sun than farther away from it. However, the basics are easy enough to understand. To begin, we can look at the the formula for finding the force of gravity, which is pretty easy. $$$F=G\frac{m_1 m_2}{r^2}$$$ Using this formula we can find the force that an object should have at orbital height.

$$F=G\frac{m_1 m_2}{r^2}$$

Where:

F is force of gravity

G is the gravitational constant, 6.67384 × 10

^{-11}N m

^{2}kg

^{-2}

m

_{1}is mass one (Earth 5.9742 × 10

^{24}kg)

m

_{2}is mass two (object)

r is the distance between the masses

Before we move on, it is important to discuss r. You may be tempted to think the distance between the Earth and an object sitting on Earth's surface is 0 (or very close). Indeed, in common usage it is. However, when dealing with gravity all distances are measured from centers of gravity. As far as these formulas are concerned everything is infinitely small with all the mass in a point (which would make everything a blackhole [which leads to an interesting consequence that blackholes behave the same as any other object at a distance beyond their event horizon]). This means that we must add the radius of Earth (6,378,100 m) to whatever height above Earth the object is at.

**Is there no gravity in orbit?**

Let's begin with a 100 kg object on Earth's surface:

$$F=6.67384 × 10^{-11} \mathrm{N\,m^2\,kg^{-2}}\frac{5.9742 × 10^{24}\,\mathrm{kg} \cdot 100\,\mathrm{kg}}{(6,378,100\,\mathrm{m}) ^2}$$

$$F=980\, \mathrm{N}$$

Newtons (N) is the SI unit of force, analogous to pounds. An apple has a weight of about one newton. It doesn't matter if you have a intuitive feel for what that force is. You just need to be able to compare it to the force in orbit, which is:

$$F=6.67384 × 10^{-11} \mathrm{N\,m^2\,kg^{-2}}\frac{5.9742 × 10^{24}\,\mathrm{kg} \cdot 100\,\mathrm{kg}}{(6,731,100\,\mathrm{m}) ^2}$$

$$F=880\, \mathrm{N}$$

Note the only difference is the increase of r from 6,378,100 m to 6,731,100 m, or a 5.5% increase. That is significant, but not overwhelming. The force on Earth's surface is 980 N while the force at ISS orbit is 880 N. This force is weight. Thus, an object in ISS orbit weighs 90% what it does on Earth's surface.

**Why don't things in orbit fall then?**

Now that we know there is only a small decrease in the force of gravity a few hundred miles up in orbit, the next question is why don't the things fall then? The answer is that they do. This isn't some silly trick answer either. Objects in orbit are falling towards Earth at all times. Indeed, orbit is accurately called free fall.

**Why don't they hit the Earth then?**

This question leads to the key aspect of orbit. It isn't the height; it's the speed. When we put something in orbit, the bulk of the fuel is used not to get the object high, but to get it moving very fast. I calculate that it takes about 9 times the energy to get an object to orbital velocity as to get it to orbital height. To understand why this is important, let's look at the picture from above again:

Now I have no idea which direction the astronaut is moving, but let's assume it's from left to right. Notice the curvature of the Earth below him. It's slight, but it's there. The Earth is nearly a perfect sphere, thus the curve is constant. If you travel in a straight line on the Earth the surface will slightly curve down and away from you. This drop off is very slow, and thus not noticed. We can figure out how fast the drop off occurs, and the answer is 7.98 inches per mile. To be clear, this means when you travel one mile on the Earth's surface you drop about 8 inches (relative to a flat plane parallel to your first step).

How long would it take to fall 8 inches? Since acceleration due to gravity is constant we can use the kinematic equations to find the answer.

$$d= v_i t + \frac{1}{2}a t^2$$

Where:

d is distance traveled (7.98 inches = 0.202692 meters)

v

_{i}is the initial velocity (0 in this case so this term drops out)

t is time, what we are looking for

a is acceleration, which in this case = g = 9.80665 m/s

^{2}

Doing some shuffling we find:

$$\frac{0.202692\,\mathrm{m}}{4.903325\,\mathrm{m/s^2}}=t^2$$

Solving for t, we find t=0.2033 seconds.

Thus, it would take about 0.2 seconds to fall 8 inches. Here's an interesting question: What would happen if you traveled a mile in the 0.2 seconds it took you to fall 8 inches? You would fall 8 inches closer to the Earth, but the surface of the Earth would also drop away about 8 inches. In the end you would be no closer or farther from the surface of the Earth. This is precisely what orbit is.

A mile in 0.2 seconds works out to about 18,000 mph. We tend to think that very fast speeds are unstable, but this is only because of atmosphere, which creates drag and slows things down. If there was no atmosphere then things would continue at whatever speed they are at indefinitely. It should now be clear why things must be lifted to be put into orbit. It is not to get them a little bit higher above Earth's center of gravity. It is to get them out of the atmosphere so that the incredible speeds required for orbit can be achieved.

Orbits tend to be ellipses, and rather complicated. But, for our purposes, we can treat them as perfect circles. Here is the formula for circular orbits:

$$v=\sqrt{\frac{m_2^2 G}{(m_1 + m_2) r}}$$

Where:

v is orbital velocity

m

_{2}is the mass of object being orbited (Earth)

G is the gravitational constant

m

_{1}is the mass of the object orbiting

r is the distance between (the center of masses of) the two objects

Notice that if m

_{1}is insignificant compared to m

_{2}that the bottom term becomes m

_{2}* r. Since the top is (m

_{2}

^{2}* G), a $$$\frac{m_2}{m_2}$$$ can drop out to 1, leaving:

$$v=\sqrt{\frac{m_2 G}{r}}$$

Since the mass of Earth is constant, the only thing that matters is height above Earth. You should already be thinking that this height won't have much effect at normal orbital heights since it is only a 5.5% increase in r. We can use Google calculator to do this math for us very easily.

At Earth's surface, v=17,685 mph

At ISS orbit, v=17,215 mph

Which comes very close to Wiki's figure of 17,239.2 mph. To put these numbers in some context, a 747's cruising speed is about 570 mph. The fastest known plane was the SR-71 with a top speed around mach 3.3 or 2200 mph.

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