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Microgravity: living on the International Space Station
Microgravity: living on the International Space Station

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3.2 Location of launch sites

You should now appreciate how difficult it is to send a rocket into space. But what about the location of the launch site on Earth? Take a look at Figure 11 and then complete Activity 5.

A 2 dimensional image of the worldwide rocket launch sites.
Figure 11 Worldwide rocket launch sites.

Activity 5 Launching a rocket

Timing: Allow approximately 15 minutes

a. 

Equator


b. 

North Pole


c. 

Atlantic Ocean


d. 

Pacific Ocean


e. 

South Pole


The correct answer is a.

Discussion

At the Equator, Earth is rotating at nearly 1700 kilometres per hour (km/h). If the rocket is launched half-way between the Equator and the North or South Pole, the speed is reduced by nearly 500 km/h. This makes it harder for the rocket to escape Earth’s gravity.

a. 

Russia and the USA


b. 

India and China


c. 

Japan and the European Space Agency


The correct answer is a.

Answer

The largest number of launch sites are in Russia and the USA.

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How can you calculate the acceleration of a rocket from the launch pad to the edge of the Earth’s atmosphere? To do this, you need to use one of Newton’s Laws. Here, you need the Second Law which relates force (F), mass (m) and acceleration (a) as shown in Equation 1.

cap f equals m times a
Equation label: Equation 1

You need to change the mass of the rocket – in this case, say 40 tonnes – into kg. So, the mass m = 40 × 1000 kg = 40 000 kg = 4 × 104 kg.

Question 4 in Activity 5 provided an approximate value of the thrust of such a rocket, that is, F = 800 000 N = 8 × 105 N.

Now, if you rearrange Equation 1 and substitute in these values, you get Equation 2.

equation sequence part 1 a equals part 2 cap f divided by m equals part 3 eight multiplication 10 super five times k times g times m postfix times s super negative two divided by four multiplication 10 super four postfix times k times g equals part 4 20 postfix times m postfix times s super negative two
Equation label: Equation 2

Here you can see that the derived units of newtons (N) have been broken down into the fundamental units of kilograms (kg), m and s. The numerical value in Equation 2 is just over twice the value of the acceleration due to gravity (g) near the surface of the Earth, which is 9.81 m/s2 (or m s–2). It means that, for every second you fall under Earth’s gravity, you travel another 9.81 m/s faster.

Now you’ve calculated how to get to the ISS. But how does it stay up in Earth’s orbit?