Download presentation

Presentation is loading. Please wait.

Published byPaloma Wolaver Modified over 6 years ago

2
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

3
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+232.43E+063.595.39E+10 Venus4.88E+246.06E+068.871.33E+11 Earth5.98E+246.38E+069.811.47E+11 Mars6.42E+233.37E+063.775.65E+10 Jupiter1.90E+276.99E+0725.953.89E+11 Saturn5.68E+265.85E+0711.081.66E+11 Uranus8.68E+252.33E+0711.081.60E+11 Neptune1.03E+262.21E+0714.072.11E+11

4
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?

5
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

6
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?

7
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)

8
G is the universal constant of gravitation, has a value of 6.67 x 10 -11 Nm 2 kg -2 Mass m 1 Mass m 2 Distance r Force F

9
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

10
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

11
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 = 6.67x10 -11 x 6.0x10 24 x 2.0x10 30 /(1.5x10 11 ) 2 F = 3.6x10 22 N

12
Take care with separation between the two masses... (r).... Consider distances between the centres and surfaces. r is the distance between their centres

Similar presentations

© 2021 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google