Page 290 - Physics Coursebook 2015 (A level)

278
Cambridge International A Level Physics
WORKED EXAMPLE
From Chapter 17, you know that the centripetal force F on a body is given by:
F = mν2 r
Consider a satellite of mass m orbiting the Earth at a distance r from the Earth’s centre at a constant speed v. Since it is the gravitational force between the Earth and the satellite which provides this centripetal force, we can write:
GMm = mν2 r2 r
where M is the mass of the Earth. (There is no need
for a minus sign here as the gravitational force and the centripetal force are both directed towards the centre of the circle.)
Rearranging gives: ν2 = GM
r
This equation allows us to calculate, for example, the speed at which a satellite must travel to stay in a circular orbit. Notice that the mass of the satellite m has cancelled out. The implication of this is that all satellites, whatever their masses, will travel at the same speed in a particular orbit. You would find this very reassuring if you were an astronaut on a space walk outside your spacecraft (Figure 18.10). You would travel at the same speed as your craft, despite the fact that your mass is a lot less than its mass. The equation above can be applied to the planets of our solar system – M becomes the mass of the Sun.
Now look at Worked example 2.
3 The Moon orbits the Earth at an average distance of 384 000 km from the centre of the Earth. Calculate its orbital speed. (The mass of the Earth is 6.0 × 1024 kg.)
Step1 Writedowntheknownquantities. r =3.84 ×108m M= 6.0×1024kg v =?
Step 2 Use the equation v 2 = GM to determine the orbital speed v. r
v2 = GM r
2
v =
6.67 × 10−11 × 6.0 × 1024 3.84×108
Figure 18.10 During this space walk, both the astronaut and the spacecraft travel through space at over 8 km s−1.
v2 =1.04×106
Hint: Don’t forget to take the square root of v2 to get v.
v =1020ms−1 ≈ 1.0×103ms−1
So the Moon travels around its orbit at a speed of
roughly 1 km s−1.
QUESTION
11 Calculate the orbital speed of an artificial satellite travelling 200 km above the Earth’s surface. (The radius of Earth is 6.4 × 106 m and its mass is 6.0 × 1024 kg.)
The orbital period
It is often more useful to consider the time taken for a complete orbit, the orbital period T. Since the distance around an orbit is equal to the circumference 2πr, it follows that:
ν = 2πr T
We can substitute this in the equation for v2. This gives:
4π2r2 = GM T2 r
and rearranging this equation gives: T2= 4π2 r3
GM
This equation shows that the orbital period T is related to
the radius r of the orbit. The square of the period is directly proportional to the cube of the radius (T 2 ∝ r3). This is