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Topic 6: Circular motion &
Gravitation
6.2 Newton’s law of gravitation
Gravitational Force & Fields
What do you know about gravity?
What does a gravitational field look like?
Can we defy gravity?
What does gravity affect?
Gravitational Force and Field
Newton proposed that a force of attraction exists
between any two masses.
This force law applies to point masses not
extended masses
However the interaction between two spherical
masses is the same as if the masses were
concentrated at the centres of the spheres.
Newton´s Law of Universal
Gravitation
Newton proposed that
“every particle of matter in the universe
attracts every other particle with a force
which is directly proportional to the product
of their masses, and inversely proportional
to the square of their distance apart”
This can be written as
F = G m1m2
r2
Where G is Newton´s constant of
Universal Gravitation
It has a value of 6.67 x 10-11
Nm2
kg-2
Calculate the gravitational pull of the
Earth on the moon
Mm =0.01230*Me
Me =5.976 x 1024
kg
R = 384400 km
Calculate the mass of the Earth
How do you determine the mass of the Earth?
Gravitational Field Strength
A mass M creates a gravitational field in
space around it.
If a mass m is placed at some point in
space around the mass M it will
experience the existance of the field in the
form of a gravitational force
We define the gravitational field strength
as the ratio of the force the mass m would
experience to the mass, M
That is the gravitational field strength at a
point, is the force exerted per unit mass on
a particle of small mass placed at that
point
The force experienced by a mass m placed a
distance r from a mass M is
F = G Mm
r2
And so the gravitational field strength of the
mass M is given by dividing both sides by m
g = G M
r2
The units of gravitational field strength are
N kg-1
The gravitational field strength is a vector
quantity whose direction is given by the
direction of the force a mass would
experience if placed at the point of interest
Field Strength at the
Surface of a Planet
If we replace the particle M with a sphere
of mass M and radius R then relying on
the fact that the sphere behaves as a point
mass situated at its centre the field
strength at the surface of the sphere will
be given by
g = G M
R2
If the sphere is the Earth then we have
g = G Me
Re
2
But the field strength is equal to the acceleration
that is produced on the mass, hence we have
that the acceleration of free fall at the surface of
the Earth, g
g = G Me
Re
2
Is g = 9.81 ?
Use the data from before and verify
whether the acceleration due to gravity
is 9.81 ms-2
Orbital speed
The speed of an object can be found from
V = 2 π r /T but can you show that
GM = v2
r
Orbital Motion
Gravitation provides the centripetal force for
circular orbital motion
The behaviour of the solar system is
summarised by Kepler´s laws
Kepler´s law state
 1. Each planet moves in an ellipse which has the
sun at one focus
 2. The line joining the sun to the moving planet
sweeps out equal areas in equal times
Deriving the Third Law
Suppose a planet of mass m moves with
speed v in a circle of radius r round the sun
of mass M
The gravitational attraction of the sun for
the planet is = G Mm
r2
From Newton’s Law of Universal
Gravitation
If this is the centripetal force keeping the
planet in orbit then
G Mm = mv2
(from centripetal equation)
r2
r
∴GM = v2
r
If T is the time for the planet to make one orbit
v = 2π r v2
= 22
π2
r2
T T2
∴ GM = 4π 2
r2
r
T2
∴ GM = 4π 2
r3
T2
∴r3
= GM
T2
4π 2
∴r3
= a constant
T2

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6.2 newtons law of gravitation

  • 1. Topic 6: Circular motion & Gravitation 6.2 Newton’s law of gravitation
  • 2. Gravitational Force & Fields What do you know about gravity? What does a gravitational field look like? Can we defy gravity? What does gravity affect?
  • 3. Gravitational Force and Field Newton proposed that a force of attraction exists between any two masses. This force law applies to point masses not extended masses However the interaction between two spherical masses is the same as if the masses were concentrated at the centres of the spheres.
  • 4. Newton´s Law of Universal Gravitation Newton proposed that “every particle of matter in the universe attracts every other particle with a force which is directly proportional to the product of their masses, and inversely proportional to the square of their distance apart”
  • 5. This can be written as F = G m1m2 r2 Where G is Newton´s constant of Universal Gravitation It has a value of 6.67 x 10-11 Nm2 kg-2
  • 6. Calculate the gravitational pull of the Earth on the moon Mm =0.01230*Me Me =5.976 x 1024 kg R = 384400 km
  • 7. Calculate the mass of the Earth How do you determine the mass of the Earth?
  • 8. Gravitational Field Strength A mass M creates a gravitational field in space around it. If a mass m is placed at some point in space around the mass M it will experience the existance of the field in the form of a gravitational force
  • 9. We define the gravitational field strength as the ratio of the force the mass m would experience to the mass, M That is the gravitational field strength at a point, is the force exerted per unit mass on a particle of small mass placed at that point
  • 10. The force experienced by a mass m placed a distance r from a mass M is F = G Mm r2 And so the gravitational field strength of the mass M is given by dividing both sides by m g = G M r2
  • 11. The units of gravitational field strength are N kg-1 The gravitational field strength is a vector quantity whose direction is given by the direction of the force a mass would experience if placed at the point of interest
  • 12. Field Strength at the Surface of a Planet If we replace the particle M with a sphere of mass M and radius R then relying on the fact that the sphere behaves as a point mass situated at its centre the field strength at the surface of the sphere will be given by g = G M R2
  • 13. If the sphere is the Earth then we have g = G Me Re 2 But the field strength is equal to the acceleration that is produced on the mass, hence we have that the acceleration of free fall at the surface of the Earth, g g = G Me Re 2
  • 14. Is g = 9.81 ? Use the data from before and verify whether the acceleration due to gravity is 9.81 ms-2
  • 15. Orbital speed The speed of an object can be found from V = 2 π r /T but can you show that GM = v2 r
  • 16. Orbital Motion Gravitation provides the centripetal force for circular orbital motion The behaviour of the solar system is summarised by Kepler´s laws Kepler´s law state  1. Each planet moves in an ellipse which has the sun at one focus  2. The line joining the sun to the moving planet sweeps out equal areas in equal times
  • 17. Deriving the Third Law Suppose a planet of mass m moves with speed v in a circle of radius r round the sun of mass M The gravitational attraction of the sun for the planet is = G Mm r2 From Newton’s Law of Universal Gravitation
  • 18. If this is the centripetal force keeping the planet in orbit then G Mm = mv2 (from centripetal equation) r2 r ∴GM = v2 r
  • 19. If T is the time for the planet to make one orbit v = 2π r v2 = 22 π2 r2 T T2 ∴ GM = 4π 2 r2 r T2 ∴ GM = 4π 2 r3 T2 ∴r3 = GM T2 4π 2 ∴r3 = a constant T2