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Published byPaloma Wolaver Modified about 1 year ago

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1.To investigate the value of g on other planets in our solar system & to practice our graph drawing and other practical skills 2.To recreate some of Newton’s work on gravitation & to establish Newton’s law of gravitation 3.To use Newton’s 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 of m/r 2 2.Next plot a good quality line graph for g against m/r 2 3.Find the Gradient of your graph Mass/kgRadius/mg N/kgm/r 2 Mercury3.18E E E+10 Venus4.88E E E+11 Earth5.98E E E+11 Mars6.42E E E+10 Jupiter1.90E E E+11 Saturn5.68E E E+11 Uranus8.68E E E+11 Neptune1.03E E E+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 of the planet squared g m g 1/r 2

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We have just “short circuited” some of Newton’s work on gravitation and for a mass of 1kg we have shown that the gravitation force F is related to the mass of and radius of the planet in the following way F m/r 2 How do we turn a proportionality into an equation we can use?

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We introduce a constant of proportionality F = Gm 1 m 2 r 2 Where F is the gravitational force between the two objects, m 1 & m 2 are the masses of the two objects & r is the separation between 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 Nm 2 kg -2 Mass m 1 Mass m 2 Distance r Force F

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Earlier we showed that the gravitational force between two objects is governed by the following F 1/r 2 This is an example of an “inverse square law”. The force is inversely proportional to the square of the separation. In practice this 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 of the Earth & sun respectively are 6.0x10 24 kg & 2.0x10 30 kg a)The diameters of the Sun & Earth respectively are 1.4x10 9 m & 1.3x10 7 m 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 sun and Earth is the same within 1% F = Gm 1 m 2 /r 2 F = 6.67x x 6.0x10 24 x 2.0x10 30 /(1.5x10 11 ) 2 F = 3.6x10 22 N

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Take care with separation between the two masses... (r).... Consider distances between the centres and surfaces. r is the distance between their centres

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