Transcript field - the SASPhysics.com

Gravitational Fields
Year 13
What is gravity?
• Gravity is a force which acts between masses.
• Its influence can be felt across free space. The
area around a mass in which it exerts a force on
another mass is known as a GRAVITATIONAL
FIELD.
• GRAVITATIONAL FIELD STRENGTH (g) is
defined as the FORCE PER UNIT MASS
(g=F/m).
• At the surface of the earth g≈9.8 N/kg, but it
varies slightly with location.
The effects of gravity
•
•
•
•
We stick to planet Earth
Existence of tides
Orbits of the moon and planets
Formation of stars…
Earth’s gravitational field strength
Kepler’s Laws
• 3 simple laws which describe
planetary motion
• Found empirically by studying
astronomical data
• Revolutionised astronomy,
overturning ideas of Aristotle
and Ptolemy
• A century later Newton
derived the laws using his
laws of motion and gravitation
Johannes Kepler
1571-1630
Kepler’s Laws
1. All planets follow elliptical orbits with the
Sun at one focus
2. A line from a planet to the Sun sweeps
out the same area per unit time interval
3.
T r
2
3
Newton’s Law of Universal
Gravitation
•“Every particle in the universe attracts
every other with a force which is
proportional to the product of their masses
and inversely proportional to the square of
their separation”.
•This is true for everything,
from subatomic particles to
stars and planets
Newton’s Law of Universal
Gravitation
1st mass
2nd mass
Attractive
force
Mm
F  G 2
r
Force
between two
masses
constant
• G = 6.67x10-11 Nm2kg-2
Separation of two
centre of masses
Inverse square law field
• Gravity is an
inverse square law
field:
• It gets weaker and
weaker as bodies
move apart, but
has infinite range
• (g→0 as r→∞)
Measuring G
• G is one of the most
difficult physical
constants to measure
accurately
• First done by Henry
Cavendish (1798)
Question
• Eratosthenes (276 - 195 B.C.) calculated the
radius of the Earth to be 6378 km. If the
acceleration due to gravity at the Earth’s surface
is 9.8 ms-2, calculate the mass of the Earth:
M em
F  mg  G 2
re
Me
so g  G 2
re


6 2
gr
9.8  6.4 10
or M e 

G
6.67 10 11
2
e
 6 10 24 kg
Example
• The mass of the earth is 18 times the mass of
Mercury, and earth is 3 times as far from the sun
as Mercury. Compare the strength of the sun’s
pull on each planet.
FMerc
FEarth
M s mM
 G
rM2
M s 18mM
M s mM
 G
 2G
2
rM2
3r M
• So the force exerted by the sun on the earth is
twice the force exerted on Mercury.
Acceleration due to gravity
• Gravitational field strength g=F/m
• Remember Newton’s 2nd law:
– F=ma
• F=ma=mg, so acceleration due to gravity =
gravitational field strength: g = 9.8 m/s2
Mm
F
M
F  G 2 , so g   G 2
r
m
r
• SO ALL BODIES FALLING UNDER GRAVITY
ACCELERATE AT THE SAME RATE (neglecting
air resistance)
Galileo’s disputed experiment
•Galileo Galilei
•1564 - 1642
•Overturned the accepted
Aristotlean idea (in force
since ~350 BC)
Journey through the centre of the
Earth
• Suppose you could drill
a hole through the Earth
and then drop into it.
How long would it take
you to pop up on the
other side of the Earth?
• So F proportional to r
– Like a mass on a spring!
• T=85 mins
Newton’s Mountain
• Imagine a cannon on a mountain top, firing
horizontally.
• What will the trajectory of the cannonball be?
• Applet
• Eventually the ball will go into orbit:
• It falls towards the earth at the same rate at
which the surface of the earth falls away from
the horizontal.
• (curvature of the earth gives ~5 m fall every
8000m)
Example
Example
• Horizontally:
• Distance=speed x
time
• =10x10=100 m
Example
1 2
s  ut  at
2
1 2
u  0, s  at
2
2
s  0.5  0.0110  0.5 m
Example
Yes!
Example
v2
g
, so v  gr
r
v  0.01 10000  10 ms -1
Satellites and Orbits
• Newton’s 1st law:
• “A body stays at rest, or if moving continues to
move with uniform velocity, unless acted on by
an external force”
• Uniform velocity means constant speed and
direction.
• So, for a body to be in orbit around another body
it must experience an ACCELERATION, caused
by a FORCE.
• This force is provided by GRAVITY.
Satellites and Orbits
•Satellites orbit larger bodies (planets or stars)
because they experience a force due to gravity
towards that body’s centre of mass.
•If the speed of the satellite is correct, it will fall
towards the surface of the body at the same rate
as the body’s surface falls away from the satellite
and a circular orbit is achieved.
•More generally, orbits are elliptical, and a circular
orbit is a special case when the two foci coincide.
Exercise
• By equating centripetal and gravitational
forces, derive an equation for the orbital
period of a satellite.
• Hence, calculate the orbital radius and
speed for a geostationary satellite (T=24
hrs)
Period of an orbit
Mm
• Setting T=24x60x60s,
2
F  G 2  m r
G=6.67x10-11Nm2kg-2,
r
24 kg
M=6x10
2

• r=42,300 km
T
• Radius of earth=6400
Mm
4 2
G 2 m 2 r
km, so r = 6.6 x earth r
r
T
3
•
V=r=3.1
km/s
r
T  2
GM
The Law of Periods
• Kepler’s 3rd Law
• The square of the period of any planet is
proportional to the cube of the semimajor
axis of its orbit.
Geostationary Orbit
• Geosynchronous orbit:
Orbital period = 1 day
• Geostationary orbit:
– A geosynchronous orbit
around the equator
– satellite appears to hover
overhead
• How would satellites in these
geosynchronous orbits appear from the
Earth?
(Resourcefulphysics.org) Not to scale
Polar orbits
Gravitational Field
• Gravity is not a contact force – it acts at a
distance
• We can represent this with field lines
• Field lines:
– Start at and end at a mass
– Never cross
Gravitational field at the surface of
the Earth
• Close to the Earth,
field strength is
approximately uniform
• Field strength:
W
M
g
G 2
m
r
• So as you rise above
the surface, g
gradually decreases
• G is the same for all
masses in the field, F
depends on the mass.
Exercise
• practice questions: field strength
Work and energy
• What is work?
– Force x distance moved in direction of force
– Amount of energy transferred from one form
to another
• If I lift an object above my head, am I
doing work?
• What has happened to the work I did on
the object?
– Stored as energy on the object – gravitational
potential energy
GPE in a constant field
• e.g. close to the Earth’s surface
• DE = F x d = mg x h
• GPE=mgh
• Assume g = 9.8 N kg-1 near the surface of the Earth.
1) Climbing a vertical rope is difficult. You have to lift your full
body weight with your arms. If your mass is 60 kg and you
climb 2.0 m, by how much do you increase your gravitational
potential energy?
2) A block of bricks is raised vertically to a bricklayer at the top
of a wall using a pulley system. If the block of bricks has a
mass of 24 kg, what is its weight? It is raised 3.0 m.
Calculate its increase in gravitational potential energy when
it reaches the top of the wall.
3) Travelling in a mountainous area, a bus of mass 3 tonnes
reaches the edge of a steep valley. There is a 1 km vertical
drop to reach the valley below, but 20 km of road to get
there. What gravitational potential energy will the bus lose in
making its descent to the valley bottom?
4) Assuming the bus in question 3 does not change its cruising
speed on the way down, where does the gravitational
potential energy go? Why is there a risk of brake failure in
this situation?
GPE in a non-uniform field
• Could we use PE=mgh for a rocket
launched to the moon?
– No, because g is not constant over such large
distances
Zero potential energy
• To move away from a mass, we must do
work to overcome gravity
– so as we move away, the GPE increases
• We choose to set the GPE at an infinite
distance from a mass to be zero (highest
possible value of GPE)
– So all other GPEs are negative
– Having defined a zero point, GPE values are
now absolute, not relative
Gravitational Potential Energy
• GPE of a mass m at a point distance r from a
(point or spherical) mass M is given by:
Mm
GPE  G
r
This minus sign NOT optional
• GPE is the energy stored as a result of a body’s
position in a gravitational field
work done moving a distance Dx :
Mm
DW  FDx  G 2 Dx
x
total work moving m from r to  :


Mm
 Mm 
W   G 2 dx  - G

x
x

r
r
Mm
 Mm 
W  0  - G
G

r 
r

Mm
GPE  0  W  G
r
GPE
Zero at ∞ distance
P.E.
Increasingly
negative as you
approach the
object providing
the field
Distance
Is it possible to escape gravity?
• Gravitational force has
infinite range, but its strength
decreases with distance.
• Gravitational P.E.=mgr
• At infinity we define the
G.P.E.=0 (it is negative at
planet surface)
• A finite amount of work is
needed to move a body to
infinity
• So if we can give a body at
the surface of the planet
sufficient K.E. to cancel out
its G.P.E. it can escape
Escape velocity
KE  PE
• Radius of the earth =
6400km
at the surface of a planet :
• g=9.8 m/s2
1 2
Mm
mv  G
• What is the escape
2
R
velocity?
v 2 GM
• v≈11.2 km/s
so

2
R
GM
• In fact this is an escape
but g  2
speed – a body with this
R
speed will escape the
2
v
Earth’s gravitational field
so
 gR or v  2 gR
2
regardless of direction
Gravitational Potential
• Gravitational Potential at a point is the
work done moving a unit mass from ∞ to
that point
• i.e. gravitational PE per unit mass
Mm
GPE  G
r
GPE
V
m
M
V  G
r
A property of the field
[Only true for a point
or spherical mass]
Equipotentials
• Lines of equal
potential (like contour
lines)
• Near the Earth’s
surface,
49.0 J/kg
V=mgh/m=gh
39.2 J/kg
• So how far apart are 29.4 J/kg
19.6 J/kg
the equipotentials
9.8 J/kg
shown here?
1m
Field strength and potential
• Like contours, if the equipotentials are close
together, the field is strong
• Field strength=-change in potential/distance
– The negative gradient of the potential
dV
g
dr