Download Higher Physics – Unit 2

Advanced Higher Physics
Unit 1
Gravitation
Gravitation investigation
Planet
Mercury
Venus
Earth
Mars
Jupiter
Saturn
Uranus
Neptune
m (kg) T (s)
r, distance
w
from sun (m) (radsˉ¹)
Fc (N)
m/r²
Planet
mass (x10^24
kg)
T(s)
w
F
r (x10^9 m)
m/r²
mercury
0.33
7600176
8.26716E-07
1.30589E+22
57.9
9.84366E-23
venus
4.87
19394640
3.23965E-07
5.53035E+22
108.2
4.15982E-22
earth
5.98
31536000
1.99238E-07
3.55123E+22
149.6
2.67201E-22
mars
0.65
59319216
1.05922E-07
1.03263E+21
141.6
3.2418E-23
jupiter
1900
3.74E+08
1.6785E-08
4.16625E+23
778.3
3.1366E-21
saturn
570
9.29E+08
6.76531E-09
3.72284E+22
1427
2.79915E-22
uranus
57
2.65E+09
2.36991E-09
9.19121E+20
2871
6.91526E-24
neptune
100
5.2E+09
1.20824E-09
6.56505E+20
4497.1
4.94464E-24
Relationship between Fg and r
Plot a graph of Fg against m/r².
Fg
(N)
m/r²(kgmˉ²)
Fg(N)
Fg against m/d^2
4.5E+23
4E+23
3.5E+23
3E+23
2.5E+23
2E+23
1.5E+23
1E+23
5E+22
0
0
1E-21
2E-21
m /d^2 (kgm ^-2)
3E-21
4E-21
Inverse square law of gravitation
The Force of gravitational attraction between any two particles
is given by:
Gm1m2
F
2
r
F is the gravitational force, measured in Newton (N)
G is the universal constant of gravitation,
G  6.67 1011 Nm2 kg 2
On data sheet
m1 and m2 are the masses of the particles in kg
r is the distance between the two particles in m
Notes:
•For orbits, r is taken as the centre of orbit which are taken
as circular. In reality, they are not circular.
•The mass of a large body can be considered to be concentrated
at its centre.
Find the mass of the sun
Find the gradient of the graph Fg against m/r².
m1
Fg  k 2
r
Gm1m2
F
2
r
However,
Therefore
k  Gm2
So mass of the sun =
with
k
m2 
G
m2= mass of the sun
G=universal constant
of gravitation
Example
A spacecraft is travelling from the Earth to the moon.
Calculate the distance from the spacecraft to the earth when the
resultant gravitational force acting on the spacecraft is 0.
(Mean radius of the moon’s orbit is 3.8 108 m )
Fmoon
Fearth
Gravitational Field Strength
The gravitational field strength reduces with increasing distance.
The gravitational field strength at a point is defined as the force
acting on a 1kg mass at that point.
F
g
m
Not in data booklet
Now we have two expressions for gravitational force:
F  mg
and
If mass of object = m=m2,
Gm1m2
F
2
r
Equating these and cancelling the mass of the object gives
Gm1
g 2
r
With m1, mass of the Earth.
Not in data booklet.
Weighing the Earth
Calculate the mass of the earth using
Gm2
g 2
r
Gravitational Fields
A gravitational field line shows the direction of the force a small test
mass would experience if placed at that point.
Isolated point mass
Two equal point masses
Gravitational Potential
The gravitational potential a point is the work done to move 1 kg
Mass from infinity to that point.
Using calculus:
V is measured in Jkgˉ¹
V=0 at infinity
Work
V
mass
Not in data booklet
 Gm
V
r
In data booklet
m mass of the planet, kg
r distance from centre of planet m
At this point, the potential is:
 Gm
V
r
r
m
The work done in bringing the satellite to
this point is:
r
 Gm
V
r
m
The potential is negative as it becomes less
when the satellite gets closer or more when
it gets further away.
Potential energy
The potential energy of a mass M at a point is:
E p  MV
Not in data booklet
Therefore:
GMm
Ep  
r
m, mass of the planet in kg
M, mass of the object in kg
r, distance from centre of planet to object in m
M
The potential energy of the satellite is:
r
GMm
Ep  
r
m
A gravitational field is a
conservative field.
The potential energy of an
object depends on its position
and not how
it was placed in that position.
Also the overall energy
required to move a mass on a
round trip is 0.
Example
A 800 kg satellite is to be placed in a circular orbit 1200 km above
the surface of the Earth.
1. Calculate the gravitational potential of the satellite.
2. Calculate its gravitational potential energy.
3. How does the gravitational energy of the satellite compare with
its gravitational energy on the Earth’s surface.
Escape velocity
Consider a mass m2, at the surface of a large
body of mass m1 and radius r.
m2
To escape the gravitational field of the
large body, the mass must be given enough
kinetic energy so that when it reaches
“infinity” its kinetic energy is 0.
r
In other words, enough kinetic energy to
overcome the potential energy at the
surface.
m1
At all points:
Ek  E p  0
1 2 GMm
 mv 
0
2
r
2GM
v 
r
2
You need to be able to derive this !
In data booklet
2GM
v
r
Black holes and photons
If the gravitational field of a body is sufficiently large, photons will
be affected by it.
In some cases the escape velocity for a body is larger than c, and
so nothing can escape its surface, not even light.
Such a body is called a black hole.
Example
Calculate the radius of the earth for it to become a black hole.
Satellites in circular motion
To maintain a circular motion,
there must be a centripetal force.
The only force which can do this is
the gravitational force.
v
m2
r
F
m1
2
m2v
Gm1m2

2
r
r
m2v 2 Gm1m2

r
r2
Gm1
v 
r
2
m1 is the mass of the large object (example Earth)
Measuring the mass of the milky way
The sun orbits the galactic center at:
r  2.6 10 m
20
v  2.2 10 ms
5
Use
1
Gm1
v 
r
2
to find the mass of the milky way.
GM
v 
r
2
Now substitute
2r
v
T
4 2 r 2 GM

2
T
r2 3
4 r
2
T 
GM
r3
 T  2
GM
Now consider energy:
GM
v 
r
2
, so
1 2 GMm
mv 
2
2r
GMm
 Ek 
2r
We already have an expression for potential energy:
GMm
Ep  
r
Therefore the total energy of the system is:
GMm
Ek  E p  
2r