Transcript Gravitation PowerPoint

Gravitation
© David Hoult 2009
© David Hoult 2009
© David Hoult 2009
© David Hoult 2009
© David Hoult 2009
© David Hoult 2009
© David Hoult 2009
F a m1m2
© David Hoult 2009
© David Hoult 2009
© David Hoult 2009
© David Hoult 2009
© David Hoult 2009
© David Hoult 2009
© David Hoult 2009
1
F a 2
r
© David Hoult 2009
F =G
m1m2
r2
where G is the universal gravitation constant
© David Hoult 2009
F =G
m1m2
r2
where G is the universal gravitation constant
N m2 kg-2
© David Hoult 2009
Testing the Inverse Square Law
of Gravitation
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© David Hoult 2009
9.8
-3 ms-2
=
2.72
×
10
602
© David Hoult 2009
9.8
-3 ms-2
=
2.72
×
10
602
v2
a=
r
© David Hoult 2009
9.8
-3 ms-2
=
2.72
×
10
602
v2
a=
r
r = 3.84 × 108 m
T = 27.3 days
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Centripetal acceleration of the moon (caused by
the force of gravity)
2.72 × 10-3 ms-2
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Conclusion
The inverse square law is a good theory
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Relation between g and G
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Relation between g and G
F
g
m
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Relation between g and G
F
g
m
Mm
FG 2
R
© David Hoult 2009
Relation between g and G
F
g
m
Mm
FG 2
R
M
gG 2
R
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we have assumed the equivalence of
inertial and gravitational mass
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Gravitational Field Strength
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The g.f.s. at a point in a gravitational field is the
force per unit mass acting on point mass
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The g.f.s. at a point in a gravitational field is the
force per unit mass acting on point mass
Units Nkg-1
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“Force per unit mass” is equivalent
to acceleration
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G.f.s. is another name for acceleration due to
gravity
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© David Hoult 2009
M
gG 2
r
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ga 1
r2
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ga 1
r2
© David Hoult 2009
© David Hoult 2009
outside the sphere
ga 1
r2
outside the sphere
ga 1
r2
© David Hoult 2009
outside the sphere
ga 1
r2
inside the sphere g a r
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outside the sphere
ga 1
r2
inside the sphere g a r
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© David Hoult 2009
© David Hoult 2009
World High Jump Record...
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World High Jump Record...
on Mars ?
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© David Hoult 2009
© David Hoult 2009
maximum height, s depends on:
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maximum height, s depends on:
initial velocity, u
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maximum height, s depends on:
initial velocity, u
acceleration due to gravity, g
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u2 = -2gs
so, for a given initial velocity
© David Hoult 2009
u2 = -2gs
so, for a given initial velocity
gs = a constant
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For a given initial velocity, the maximum height
reached by the body is inversely proportional to the
acceleration due to gravity
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s a 1
g
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s a 1
g
sg = a constant
© David Hoult 2009
s a 1
g
gs = a constant
g1s1 = g2s2
or
s1
g2
=
s2
g1
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Gravitational Potential
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The potential at a point in a gravitational field is
the work done per unit mass moving point mass
from infinity to that point
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The potential at a point in a gravitational field is
the work done per unit mass moving point mass
from infinity to that point
units of potential J kg-1
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© David Hoult 2009
© David Hoult 2009
© David Hoult 2009
w = Fs but in this situation the force is not of
constant magnitude
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w = Fs but in this situation the force is not of
constant magnitude
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It is clear that the work done will depend on:
© David Hoult 2009
It is clear that the work done will depend on:
 the mass of the planet, M
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It is clear that the work done will depend on:
 the mass of the planet, M
 the distance, r of point p from the planet
© David Hoult 2009
It is clear that the work done will depend on:
 the mass of the planet, M
guess: w a M
 the distance, r of point p from the planet
© David Hoult 2009
It is clear that the work done will depend on:
 the mass of the planet, M
guess: w a M
 the distance, r of point p from the planet
guess: w a 1/r
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...it can be shown that...
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GM
w =
r
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A body at infinity, has zero gravitational potential
“at infinity” means that the body is out of the
gravitational field
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A body at infinity, has zero gravitational potential
“at infinity” means that the body is out of the
gravitational field
All bodies fall to their lowest state of potential
(energy)
© David Hoult 2009
A body at infinity, has zero gravitational potential
“at infinity” means that the body is out of the
gravitational field
All bodies fall to their lowest state of potential
(energy)
All gravitational potentials are therefore negative
quantities
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V =
GM
r
© David Hoult 2009
V =
GM
r
Therefore the gravitational potential energy possessed by a
body of mass m placed at point p is given by
© David Hoult 2009
V =
GM
r
The gravitational potential energy possessed by a body of
mass m placed at point p is given by
GPE = Vm
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Escape Velocity
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© David Hoult 2009
© David Hoult 2009
© David Hoult 2009
© David Hoult 2009
G P E = zero
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G P E = zero
To find the minimum velocity, ve which will cause the rocket
to escape the Earth’s gravity, assume K E of distant rocket
is also equal to zero.
© David Hoult 2009
G P E = zero
To find the minimum velocity, ve which will cause the rocket
to escape the Earth’s gravity, assume K E of distant rocket
is also equal to zero.
As the body is moving away from the planet it is losing
K E and gaining G P E
© David Hoult 2009
G P E = zero
To find the minimum velocity, ve which will cause the rocket
to escape the Earth’s gravity, assume K E of distant rocket
is also equal to zero.
As the body is moving away from the planet it is losing
K E and gaining G P E
DKE = DGPE
© David Hoult 2009
If the mass of the rocket is m, then the G P E it possesses
at the surface of the planet is
GPE =
GMm
R
© David Hoult 2009
If the mass of the rocket is m, then the G P E it possesses
at the surface of the planet is
GPE =
GMm
R
DGPE =
GMm
r
© David Hoult 2009
If the mass of the rocket is m, then the G P E it possesses
at the surface of the planet is
GPE =
GMm
R
DGPE =
GMm
R
D K E = ½mve2
© David Hoult 2009
½mve
2
=
GMm
R
© David Hoult 2009
2GM
ve 
R
Also, as g = GM/R2
© David Hoult 2009
2GM
ve 
R
Also, as g = GM/R2
ve  2gR
© David Hoult 2009