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F = G m o m e r 2

and the acceleration of an object of mass m o (in terms of the force acting on it) is

a o = F m o

So we substitute equation [link] into Equation  [link] , and we find that

a o = G m e r 2

Since it doesn't matter what m o is, this tells us that the acceleration on a body (due to the Earth's gravity) does not depend on the mass of the body. Thus all objects experience the same gravitational acceleration. The force on different bodies will be different but the acceleration will be the same. Due to the fact that this acceleration caused by gravity is the same on all objects we label it differently, instead of using a we use g which we call the gravitational acceleration.

Comparative problems

Comparative problems involve calculation of something in terms of something else that we know. For example, if you weigh 490 N on Earth and the gravitational acceleration on Venus is 0,903 that of the gravitational acceleration on the Earth, then you would weigh 0,903 x 490 N = 442,5 N on Venus.

Principles for answering comparative problems

  • Write out equations and calculate all quantities for the given situation
  • Write out all relationships between variable from first and second case
  • Write out second case
  • Substitute all first case variables into second case
  • Write second case in terms of first case

A man has a mass of 70 kg. The planet Zirgon is the same size as the Earth but has twice the mass of the Earth. What would the man weigh on Zirgon, if the gravitational acceleration on Earth is 9,8 m · s - 2 ?

  1. The following has been provided:

    • the mass of the man, m
    • the mass of the planet Zirgon ( m Z ) in terms of the mass of the Earth ( m E ), m Z = 2 m E
    • the radius of the planet Zirgon ( r Z ) in terms of the radius of the Earth ( r E ), r Z = r E
  2. We are required to determine the man's weight on Zirgon ( w Z ). We can do this by using:

    w = m g = G m 1 · m 2 r 2

    to calculate the weight of the man on Earth and then use this value to determine the weight of the man on Zirgon.

  3. w E = m g E = G m E · m r E 2 = ( 70 kg ) ( 9 , 8 m · s - 2 ) = 686 N
  4. Write the equation for the gravitational force on Zirgon and then substitute the values for m Z and r Z , in terms of the values for the Earth.

    w Z = m g Z = G m Z · m r Z 2 = G 2 m E · m r E 2 = 2 ( G m E · m r E 2 ) = 2 w E = 2 ( 686 N ) = 1 372 N
  5. The man weighs 1 372 N on Zirgon.

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A man has a mass of 70 kg. On the planet Beeble how much will he weigh if Beeble has mass half of that of the Earth and a radius one quarter that of the Earth. Gravitational acceleration on Earth is 9,8 m · s - 2 .

  1. The following has been provided:

    • the mass of the man on Earth, m
    • the mass of the planet Beeble ( m B ) in terms of the mass of the Earth ( m E ), m B = 1 2 m E
    • the radius of the planet Beeble ( r B ) in terms of the radius of the Earth ( r E ), r B = 1 4 r E
  2. We are required to determine the man's weight on Beeble ( w B ). We can do this by using:

    w = m g = G m 1 · m 2 r 2

    to calculate the weight of the man on Earth and then use this value to determine the weight of the man on Beeble.

  3. w E = m g E = G m E · m r E 2 = ( 70 kg ) ( 9 , 8 m · s - 2 ) = 686 N
  4. Write the equation for the gravitational force on Beeble and then substitute the values for m B and r B , in terms of the values for the Earth.

    w B = m g B = G m B · m r B 2 = G 1 2 m E · m ( 1 4 r E ) 2 = 8 ( G m E · m r E 2 ) = 8 w E = 8 ( 686 N ) = 5 488 N
  5. The man weighs 5 488 N on Beeble.

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Physics is all about being simple - all we do is look at the world around us and notice how it really works. It is the one thing everyone is qualified to do - we spend most of our time when we are really young experimenting to find out how things work.

The actual force of air resistance is quite complicated. Experiment by moving a book through the air with the face of the book and then the side of the book forward, you will agree that the area of the book makes a difference as to how much you must work in order to move the book at the same speed in both cases. This is why racing cars are slim-lined in design, and not shaped like a big box!

Exercise

  1. Two objects of mass 2m and 3m respectively exert a force F on each other when they are a certain distance apart. What will be the force between two objects situated the same distance apart but having a mass of 5m and 6m respectively?
    1. 0,2 F
    2. 1,2 F
    3. 2,2 F
    4. 5 F
  2. As the distance of an object above the surface of the Earth is greatly increased, the weight of the object would
    1. increase
    2. decrease
    3. increase and then suddenly decrease
    4. remain the same
  3. A satellite circles around the Earth at a height where the gravitational force is a factor 4 less than at the surface of the Earth. If the Earth's radius is R, then the height of the satellite above the surface is:
    1. R
    2. 2 R
    3. 4 R
    4. 16 R
  4. A satellite experiences a force F when at the surface of the Earth. What will be the force on the satellite if it orbits at a height equal to the diameter of the Earth:
    1. 1 F
    2. 1 2 F
    3. 1 3 F
    4. 1 9 F
  5. The weight of a rock lying on surface of the Moon is W. The radius of the Moon is R. On planet Alpha, the same rock has weight 8W. If the radius of planet Alpha is half that of the Moon, and the mass of the Moon is M, then the mass, in kg, of planet Alpha is:
    1. M 2
    2. M 4
    3. 2 M
    4. 4 M
  6. Consider the symbols of the two physical quantities g and G used in Physics.
    1. Name the physical quantities represented by g and G .
    2. Derive a formula for calculating g near the Earth's surface using Newton's Law of Universal Gravitation. M and R represent the mass and radius of the Earth respectively.
  7. Two spheres of mass 800g and 500g respectively are situated so that their centers are 200 cm apart. Calculate the gravitational force between them.
  8. Two spheres of mass 2 kg and 3 kg respectively are situated so that the gravitational force between them is 2,5 x 10 - 8 N. Calculate the distance between them.
  9. Two identical spheres are placed 10 cm apart. A force of 1,6675 x 10 - 9 N exists between them. Find the masses of the spheres.
  10. Halley's comet, of approximate mass 1 x 10 15 kg was 1,3 x 10 8 km from the Earth, at its point of closest approach during its last sighting in 1986.
    1. Name the force through which the Earth and the comet interact.
    2. Is the magnitude of the force experienced by the comet the same, greater than or less than the force experienced by the Earth? Explain.
    3. Does the acceleration of the comet increase, decrease or remain the same as it moves closer to the Earth? Explain.
    4. If the mass of the Earth is 6 x 10 24 kg, calculate the magnitude of the force exerted by the Earth on Halley's comet at its point of closest approach.

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Source:  OpenStax, Siyavula textbooks: grade 11 physical science. OpenStax CNX. Jul 29, 2011 Download for free at http://cnx.org/content/col11241/1.2
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