Page 292 - Physics Coursebook 2015 (A level)

Cambridge International A Level Physics
280
Summary
■■ The force of gravity is an attractive force between any two objects due to their masses.
■■ The gravitational field strength g at a point is the gravitational force exerted per unit mass on a small object placed at that point – that is:
g = mF
■■ The external field of a uniform spherical mass is the same as that of an equal point mass at the centre of the sphere.
■■ Newton’s law of gravitation states that:
Any two point masses attract each other with a force that is directly proportional to the product of their masses and inversely proportional to the square of their separation.
■■ The equation for Newton’s law of gravitation is:
F = GMm r2
For a satellite to stay above a fixed point on the equator, it must take exactly 24 hours to complete one orbit (Figure 18.12). We know:
G = 6.67×10−11 Nm2 kg−2
T = 24 hours = 86400s
M = 6.0 × 1024 kg
r3 = GMT2 = 6.67×10−11 ×6.0×1024 ×(86400)2 4π2 4π2
QUESTIONS
= 7.66 × 1022 m3
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r= 7.66 × 1022 ≈ 4.23 × 107 m
So, for a satellite to occupy a geostationary orbit, it must be at a distance of 42 300 km from the centre of the Earth and at a point directly above the equator. Note that the radius of the Earth is 6400 km, so the orbital radius is 6.6 Earth radii from the centre of the Earth (or 5.6 Earth radii from its surface). Figure 18.12 has been drawn to give an impression of the size of the orbit.
Clarke belt
Figure 18.12 Geostationary satellites are parked in the ‘Clarke belt’, high above the equator. This is a perspective view; the Clarke belt is circular.
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For any future mission to Mars, it would be desirable to set up a system of three or four geostationary (or ‘martostationary’) satellites around Mars to allow communication between the planet and Earth. Calculate the radius of a suitable orbit around Mars.
Mars has mass 6.4 × 1023 kg and a rotational period of 24.6 hours.
Although some international telephone signals are sent via satellites in geostationary orbits, most are sent along cables on the Earth’s surface. This reduces the time delay between sending and receiving the signal. Estimate this time delay for communication via a satellite, and explain why it is less significant when cables are used.
You will need the following:
• radius of geostationary orbit = 42 300 km
• radius of Earth = 6400 km
• speed of electromagnetic waves in free space c=3.0×108ms−1