*Before you read this, I suggest you read* posts 16.16 and 17.13

Why does a satellite stay in orbit around the earth? If we think of something going round in a circle (like the hammer in post 17. 13 before it is thrown), it appears that something must be pulling it towards the earth. What can this “something” be? The only possibility is the force exerted by the earth’s gravitational field (post 16.16) that has a magnitude of *GMm*/*r*^{2} where *G* is the universal gravitational constant, *M* is the mass of the earth, *m* is the mass of the satellite and *r* is the distance of the satellite from the centre of the earth.

This gravitational force provides the centripetal force (post 17.13) to keep the satellite moving in a circle. The magnitude of the centripetal force for an object of mass *m* and angular speed (post 17.12) *ω* is *m**ω*^{2}*r* where *r* is the radius of the circle (post 17.13). Since this force is provided by the earth’s gravitational field, we can write the equation

*m**ω*^{2}*r* = *GMm*/*r*^{2}.

If we divide both sides of this equation by *m* and then divide each side by *r*, we get the result that

*ω*^{2} = *GM*/*r ^{3}*. (1)

Dividing both sides of equation 1 by *ω*^{2} and multiplying both sides by *r ^{3}* gives the result that

*r*^{3} = (*GM*/*ω*^{2}). (2)

To find *r*, we take the *cube root* of both sides of equation 2. The cube root of a number is the number that we need to multiply by itself three times to get the original number. So, the cube root of 8 is 2 because 2 × 2 × 2 = 8. We could write this as 8^{1/3} = 2. Taking the cube root of both sides of equation 3 gives

*r* = (*GM*/*ω*^{2})^{1/3}. (3)

Equation 3 relates the angular speed of the satellite to the radius of its orbit.

As an example, let’s find the radius for the orbit of a geostationary satellite. Let’s suppose that we always want the satellite to appear in the same place when viewed from a fixed point on the equator. The satellite must have an angular speed of 2π/(24 × 60 × 60) = 7.272 × 10^{-5} rad.s^{-1} (post 17.12) because it rotates through 2π radians (1 complete revolution, see post 17.11) every 24 hours or 24 × 60 × 60 seconds. (If you’re not sure about the way I’ve written the result for the angular speed, see post 16.7)

*G* has a value of 6.674 ×10^{-11} m^{3}.kg^{-1}.s^{-2} and *M* has a value of 5.972 ×10^{24} kg. Putting these values and the required value for *ω* into equation 3 gives the result that *r* = 4.2 × 10^{7} m which equals 4.2 × 10^{4} km. (If you want to repeat the calculation and aren’t sure how to work out the cube root, see the appendix).

Since the radius of the earth is 0.6 × 10^{4} km, the height of the satellite above the surface of the earth is (4.2 – 0.6) × 10^{4} = 3.6 × 10^{4} km.

To get the satellite into orbit it is first launched into space using a rocket (see posts 17.26 and 17.27). When it has reached the required height, it is released and propelled at right angles to the direction of the rocket with a speed *v* = *ωr* (see post 17.12), where *ω* and *r* have the values we have already calculated, in the direction of the earth’s rotation (see diagram above).

Why doesn’t a satellite fall? It does! It is in orbit because it is being attracted by the earth’s gravitational field – which is another way of saying that it’s falling (posts 16.16 and 17.20) If it didn’t fall (because the gravitational field suddenly ceased to exist) it would fly off at a tangent to its orbit – just like the hammer of post 17.13 when the thrower stops pulling the cable.

But sometimes satellites do fall closer to earth. When they meet the earth’s atmosphere, the drag forces lead to loss of a lot of energy (post 17.17). This energy is lost as heat (post 16.35) and the satellite debris burns – the result can sometimes be seen in the night sky. Why do they fall closer to earth? One reason is because of collisions with other objects.

But the important point is that a satellite in stable orbit is always falling. The gravitational force that makes it fall provides the centripetal force to keep it in orbit.

*Related posts*

17.27 Getting into space

17.26 Rockets

17.16 Why do planes fly?

*Appendix*

It is easy to calculate that *r*^{3} = 7.537 × 10^{22} m^{3}, using equation 2. But how do we calculate *r* from this result?

Type **=** **log10(7.537*10^22)** into a cell in the Excel spreadsheet. You will get the answer **22.877198**. This answer means that 10^{22.877198} = 7.537 × 10^{22}.

So, the cube root of 7.537 × 10^{22} is equal to 10^{22.877198/3} = 10^{7.6257328}.

Now type **= 10^7.6257328** into another cell in Excel to obtain the value of *r*, written in the usual way, in m.