1. NCEA Level 3 Physics
Gravitational Fields

2. Intro- Force Fields
The concept of a field is a valuable ‘thinking model’ to help scientists (like yourselves) to explain the interaction between objects which are not in direct contact.

3. Check out the demonstrations!!
Types of fields
Electrical- due to charged objects
Magnetic- due to moving charges
Gravitational- due to masses
Check out the demonstrations!!
Static moving coke can
Electromagnet
Slinky

4. Gravitational Fields Newton’s falling apple
There is a gravitational attraction between any two objects with mass- this force is usually too weak to be noticed unless one of the objects considered has a massive amount of mass (or just very large!).

5. Curved Space
Have you ever visited the planetarium and witnessed a gravity well model in action?  There it stands before you, shaped like a huge vortex roughly six feet in diameter.  There are usually one inch diameter steel balls rolling in circles of ever decreasing radius as their energy is lost to friction.  Ultimately they spiral towards the center of the model where they are gobbled up by the well and magically recycled for their next simulated trajectory.  Some planetariums display the balls rolling in elliptical orbits, parabolic trajectories, and near circular orbits.  Usually the elliptical orbit experiences precession (or rotation) of the ellipse itself, ultimately becoming near circular shortly before the ball disappears down the well.  There is more to this museum display than meets the eye, and we shall explore this on SpaceAnimations.org. 
Many of the phenomena observed in planetary motion may be represented using a surface of revolution and a rolling ball.  The ball plays the role of the satellite or planet, while the surface represents the gravitational field.  If the surface is properly contoured, the ensuing motion of a ball projected along it will reproduce the motion of the planets.

6. Gravitational Force Field
Force Field (Gravitational Field)- any object placed within the field will experience a force
Any object with mass creates a field around it-
Field lines- the radial lines around a sphere (planet) show the direction a small mass would follow if in the field
The density of the field lines corresponds with the field strength

7. Why does it have to be small?
By definition
Gravitational Field Strength, g, is the force per unit mass on a small test mass placed in a field
g = F / m
Nkg-1 N / kg
Where F is the Force acting on a mass, m in the field.
Why does it have to be small?

8. …eh because
A small mass does not have a significant force field of its own so it will not affect the field which we are concerned with.

9. …just so you know Field lines close to the Earth look like this-
This is an example of a uniform field where the strength is the same everywhere- we use this when distances are considered to be relatively small (couple of thousand miles).

10. Black Hole Fun!
Physics Ripped apart at the seems… where all you’ve been studying falls to pieces and no-one can save you!!
The average sized black hole is 8 times more massive than our own Sun all concentrated into a space no larger than your finger nail

11. Field strength on a point mass
Consider Newton’s Laws on the point mass (2 ways of expressing F)-
F = Gm1m2 (Law of gravitation)
F = m2 g (Newton’s 2nd Law- F = m a)
Equate
m2 g = Gm1m2 r2 r2

12. Field strength on a point mass
g = Gm
m is the mass of the object causing the gravitational field
r is the distance from the centre of the mass m to the point mass
G is the universal constant of gravity r2

13. Brings us back… Gravity constant
g = Gm = F m
g also follows an inverse square law
Draw a graph to show the relationship between g and r r2 g r

14. Inverse square Law Field strength g, weakens as distance r, increases12 22 32

15. Calculating the Mass of the Earth
Value of g on the Earth’s surface
g = 9.81 Nkg-1
Radius of the Earth
r = 6400km = 6.4x106m
Using the field strength equation-
g = Gm
Rearrange for mass- m = g r2 G
= 9.81 x (6.4x106m)2 / 6.67x10-11
6.02 x 1024 kg r2

16. Did you Know… (probably not)
The mass of the Earth increases every year because of 3,000 tonnes of meteorite debris that hits its surface from space.

17. Get you thinking…
Using the new formula and Newton’s second law of motion, can you show that the value for the gravitational field strength is also the value for acceleration due to free-fall?

18. Fields- syllabus reference
10 Gravitational force between point masses (Newton’s law)
- Knowledge and use of F = Gm1m2/r2.
- G introduced as constant with units.
11 Gravitational field of a point mass
- Knowledge and use of g = Gm/r2.
- Calculation of the Earth’s mass.

19. Well.. It wasn’t as simple as that!
Newton’s Law
In 1689 an apple fell…
Well.. It wasn’t as simple as that!

20. Newton’s Law
If the force of gravity reaches to the top of the highest tree, might it not reach even further; in particular, might it not reach all the way to the orbit of the Moon?

21. Just like the acceleration of the apple…
The acceleration due to gravity changes the velocity of the Moon so that it follows an orbit around the Earth

22. Sent into Orbit!
At just the right speed and trajectory the canon ball will fall towards the Earth due to the gravitational force at the same rate as the Earth’s surface curves away from it.
The canon ball has been sent into orbit!

23. Newton’s Law of Gravitation
Every object attracts every other object along a line that joins their centres. It is proportional to the product of their masses and inversely proportional to the square of their distances apart

24. The equation r2 F = Gm1m2 F is the gravitational force of attraction
m1m2 is the product of the two masses
r is the distance between the centre of the masses
G is the gravitational constant with a value of 6.67 x 10-11 r2

25. Wee Example
The Moon’s orbit of the Earth is at a radius 3.84x108m, the mass of the moon is 7.35x1022kg and the Earth’s mass is 6.00x1024kg
Calculate the size of the force which keeps the Moon in orbit of the Earth.
F = Gm1m = x10-11 x 6.00x1024 x 7.35x1022
= 1.99x1020 N
r2 (3.84x108m)2

26. And what about this rocket…
A rocket of mass kg is fired towards the Moon. What is the net gravitational force on the rocket when it is 3.00x108m away from the centre of the Earth?
FE = 187N
FM = 29.2N
Net Force =187 – 29.2 = 158N
(Towards the Earth!)

27. How attractive are you?
Use the equation to approximately work out the force of attraction you have with the person closest to you
Why is the value so small?

28. An inverse square relationship
As the distance between two objects increases the attractive force decreases (inverse) by a directly square proportion-

29. Comet Action

30. So…
If the force between two objects is 200N when they are 5m apart, what is the force between them when they are-
A) 10m apart? (2 times the distance)
B) 20m apart? (4 times the distance)
C) 30m apart? (6 times the distance)
200/4 = 50N /16= 12.5N 200/36= 5.56N

31. The unit of G (G unit)
Re-arrange the N. Law of G. to make G the subject-
G = F r2 / m1m2
Unit of G is therefore = N m2 kg-2

32. Some extra notes to be aware of…
Gravitational force is equal and opposite for two masses (each ‘pull’ each other with the same force)
The masses are considered to be point masses with all their mass concentrated at their centre
Gravitational forces are very weak unless we are looking at enormous masses

33. For our previous rocket…
At what distance from the Earth is their no resultant forces acting on the rocket?
Force due to Earth = Force due to Moon
GmEmR = GmMmR
(3.84x108m- R)2 R2
3.84x108m
Moon
Earth R

34. For our previous rocket…
GmEmR = GmMmR
mE x R2 = mM x (3.84x108m- R)2
(3.84x108m- R)2 R2
Solve using quadratic equation…
3.84x108m
Moon
Earth R

35. Or… x
GmEmR = GmMmR
mM = mE
= mE mM
Moon
Earth y X2 Y2 X2 Y2 X2 Y2

36. Take the square root… = 9 One TENTH of 3.84x108 is 0.384x108
= 9
X is 9/10 of total distance X
Y is 1/10 of total distance Y
One TENTH of 3.84x108 is 0.384x108
Therefore, distance from the Earth where the forces are balanced is 3.84x108 – 0.384x108
= 3.456x108m

37. Syllabus Reference 11 Gravitational field of a point mass
- Knowledge and use of g = Gm/r2.
- Calculation of the Earth’s mass.

38. Bit of Homewrok
Try the past paper question…

39. Syllabus Reference 12 Planetary and satellite motion
Quantitative treatment of circular orbits only.
A formal statement of Kepler’s laws is not required, but candidates should be able to show that the mathematical form of the third law (t2 proportional to r3) is consistent with Newton’s law of gravitation.
Geostationary satellites.

40. The force required to keep a planet in circular orbit
The two equations for
F = Gm1m2
F = m v2 1
Newton’s Law for Gravitation r2 2
Centripetal force 2 r
The force required to keep a planet in circular orbit 2

41. The force required to keep a planet in circular orbit
Massive object!
The force required to keep a planet in circular orbit 1
Acceleration towards centre 2

42. Sub into equation
g = Gm / r2 2 1
m g = 2 2 v2 g r

43. Geostationary Orbits
Some satellites orbit exactly over the Earth's equator and make one orbit per day.
- Period of orbit T = 24hrs
Since the Earth rotates once on its axis per day, the satellite seems to hover over the same spot on Earth all the time.

44. Satellites in Orbit
The first satellite to be put into Geostationary orbit was way back in 1963
Since then there are now over 600 at a height of 35,000km above the Earth
What would they be used for?
Any idea what the satellite ‘Hotbird 1’ is tracking?

45. Kepler’s Third Law… you gotta know this
The ratio of the period of a mass’ orbit squared (T2) to the mean radius of it’s orbit cubed (R3) is the same constant value k for all the planets which orbit the Sun.
Planet
Period
(s)
Average
Dist. (m)
T2/R3
(s2/m3)
Earth
3.156 x 107 s
x 1011
2.977 x 10-19
Mars
5.93 x 107 s
2.278 x 1011
2.975 x 10-19

46. What’s more
Newton’s Law of gravitation is consistent with Kepler’s 3rd Law
- Consider a planet with mass mp to orbit in nearly circular motion about the sun of mass mS
The Net Centripetal Force F of the orbiting planet p
The Net Centripetal Force F of the gravitational field

47. Let’s do some substituting and re-arranging…
v = 2 π r / T v2 = 4 π2 r2 / T2 p =
Gms
4 π2 r2 = r T2
Gms
4 π2
Divide each side by r2 =
This value will be the same for all planets r3 T2 T2
Gms
Multiply each side by T2 =
4 π2 r3 T2
4 π2
Divide each side by Gms = r3
Gms

48. What’s more
Newton’s Law of gravitation is consistent with Kepler’s 3rd Law
- Consider a planet with mass mp to orbit in nearly circular motion about the sun of mass mS
The Net Centripetal Force F of the orbiting planet p
The Net Centripetal Force F of the gravitational field

49. Question 1 (Jan 2001)
The planet Mars has radius 3.39x106m and mass 6.50x1023kg. The length of a day on Mars is 8.86x104s (24.6 days)
a) At what height above the surface of Mars should a geostationary satellite be placed?
b) Calculate the acceleration of free-fall on the surface of Mars

50. Mars has two moons, Phobos (radius of orbit 9. 38x103km, 0
Mars has two moons, Phobos (radius of orbit 9.38x103km, days) and Deimos (radius of orbit 23.5x103km)
c) What is the period of Deimos’ orbit?
d) What are the three features of a geostationary satellite?

51. Question 2 (Jan 2002) State in words, Newton’s law of Gravitation
The period T of a satellite around a planet is related to its radius r by
where M is the mass of the planet being orbited. Show this is consistent with Newton’s Law of Gravitation
4 π2 r3 T2 =
Gms

52. Using the data radius of Earth 6. 37x106m and mass 5
Using the data radius of Earth 6.37x106m and mass 5.98x1024kg, a satellite is in geostationary orbit around the Earth.
c) What is meant by a geostationary orbit?
d) Calculate the height of the satellite above the Earth’s surface.

53. Circular Motion What is angular velocity (illustration?)
What is linear velocity?
Define frequency and period and show how they are related to angular velocity
Centripetal acceleration and hence force (with directions…)
Again link in period, velocity and angular velocity

54. Newton’s Law of gravitation
State in words
What does G stand for, what’s special about it?
Show in diagram form
Value of g, equation for calculations
Inverse square relationship

55. Geostationary Orbits Meaning of a Geostationary Orbit
Newton’s second law for a circular orbit in terms of angular velocity and period T
Equate Newton’s two equations
Show how Kepler’s Law is consistent with Newton’s Law