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1. To investigate the value of gon other planets
in our solar system & to practice our graph
drawing and other practical skills
2. To recreate some of Newtons work on
gravitation & to establish Newtons law of
gravitation
3. To use Newtons law to establish the
gravitational force of attraction between two
objects
Book Reference : Pages 59-61
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Using the supplied planetary data complete the following:
1. For each planet complete the table by calculating the value ofm/r2
2. Next plot a good quality line graph for g against m/r2
3. Find the Gradient of your graph
Mass/kg Radius/m g N/kg m/r2
Mercury 3.18E+23 2.43E+06 3.59 5.39E+10
Venus 4.88E+24 6.06E+06 8.87 1.33E+11
Earth 5.98E+24 6.38E+06 9.81 1.47E+11
Mars 6.42E+23 3.37E+06 3.77 5.65E+10
Jupiter 1.90E+27 6.99E+07 25.95 3.89E+11
Saturn 5.68E+26 5.85E+07 11.08 1.66E+11
Uranus 8.68E+25 2.33E+07 11.08 1.60E+11
Neptune 1.03E+26 2.21E+07 14.07 2.11E+11
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You should have a straight line graph. What does
this tell us about
The relationship between g and the mass of
the planet?
The relationship between g and the radius of
the planet?
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The straight line graph tells us that
g is directly proportional to the mass of
the planet
g is inversely proportional to the radius ofthe planet squared
g m
g 1/r2
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We have just short circuited some of Newtons
work on gravitation and for a mass of 1kg wehave shown that the gravitation force F is related
to the mass of and radius of the planet in the
following way
F m/r2
How do we turn a proportionality into anequation we can use?
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We introduce a constant of proportionality
F = Gm1m2
r2
Where Fis the gravitational force between
the two objects, m1& m2are the masses of
the two objects & ris the separationbetween the centres of the masses (think
point masses)
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G is the universal constant of gravitation,
has a value of 6.67 x 10-11Nm2kg-2
Mass m1 Mass m2
Distance r
Force F Force F
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Earlier we showed that the gravitational
force between two objects is governed bythe following
F 1/r2
This is an example of an inverse square
law. The force is inversely proportional to
the square of the separation. In practicethis means that the force reduces quickly as
r increases, we will see many inverse square
laws
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The distance from the centre of the sun to the
centre of the Earth is 1.5x10
11
m & the masses ofthe Earth & sun respectively are 6.0x1024kg &
2.0x1030kg
a) The diameters of the Sun & Earthrespectively are 1.4x109m & 1.3x107m why
is it reasonable to consider them both to be
point masses?
b) Calculate the force of gravitational
attraction between the Earth & the Sun
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The diameters of both are small compared to the
separation. Distance between any part of the sunand Earth is the same within 1%
F = Gm1m2/r2
F = 6.67x10-11x 6.0x1024x 2.0x1030/(1.5x1011)2
F = 3.6x1022N
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Take care with separation between the two
masses... (r).... Consider distances between thecentres and surfaces.
r is the distance between their centres