NCEA Level 3 Physics Gravitational Fields Intro Force

NCEA Level 3 Physics Gravitational Fields

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.

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

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!).

Curved Space

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

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

…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.

…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).

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

Field strength on a point mass m 1 Consider Newton’s Laws on the point mass (2 ways of expressing F) • F = Gm 1 m 2 (Law of gravitation) r 2 • F = m 2 g (Newton’s 2 nd Law- F = m a) Equate m 2 g = Gm 1 m 2 r 2 m 2

Field strength on a point mass g= Gm r 2 • 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

Brings us back… Gravity constant g= Gm r 2 = F m g also follows an inverse square law Draw a graph to show the relationship between g and r g

Inverse square Law • Field strength g, weakens as distance r, increases 12 22 32

Calculating the Mass of the Earth • Value of g on the Earth’s surface g = 9. 81 Nkg-1 • Radius of the Earth r = 6400 km = 6. 4 x 106 m Using the field strength equationg = Gm r 2 Rearrange for mass= m = g r 2 G 9. 81 x (6. 4 x 106 m)2 / 6. 67 x 10 -11 6. 02 x 1024 kg

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

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?

Fields- syllabus reference 10 Gravitational force between point masses (Newton’s law) - Knowledge and use of F = Gm 1 m 2/r 2. - G introduced as constant with units. 11 Gravitational field of a point mass - Knowledge and use of g = Gm/r 2. - Calculation of the Earth’s mass.

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

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?

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

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!

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

The equation F = Gm 1 m 2 r 2 F is the gravitational force of attraction m 1 m 2 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

Wee Example • The Moon’s orbit of the Earth is at a radius 3. 84 x 108 m, the mass of the moon is 7. 35 x 1022 kg and the Earth’s mass is 6. 00 x 1024 kg • Calculate the size of the force which keeps the Moon in orbit of the Earth. F = Gm 1 m 2 r 2 = 1. 99 x 1020 N = 6. 67 x 10 -11 x 6. 00 x 1024 x 7. 35 x 1022 (3. 84 x 108 m)2

And what about this rocket… • A rocket of mass 42 000 kg is fired towards the Moon. What is the net gravitational force on the rocket when it is 3. 00 x 108 m away from the centre of the Earth? • FE = 187 N • FM = 29. 2 N • Net Force =187 – 29. 2 = 158 N (Towards the Earth!)

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?

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

Comet Action

So… • If the force between two objects is 200 N when they are 5 m apart, what is the force between them when they are • A) 10 m apart? (2 times the distance) • B) 20 m apart? (4 times the distance) • C) 30 m apart? (6 times the distance) 200/4 = 50 N 200/16= 12. 5 N 200/36= 5. 56 N

The unit of G (G unit) • Re-arrange the N. Law of G. to make G the subject. G = F r 2 / m 1 m 2 Unit of G is therefore = N m 2 kg-2

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

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 Gm. Em. R = (3. 84 x 108 m- R)2 Gm. Mm. R R 2 3. 84 x 108 m Eart h R Moon

For our previous rocket… Gm. Em. R = (3. 84 x 108 m- R)2 m. E x R 2 Gm. Mm. R R 2 = m. M x (3. 84 x 108 m- R)2 Solve using quadratic equation… 3. 84 x 108 m Eart h Moon R

Or… x Eart h y Gm. Em. R = Gm. Mm. R X 2 Y 2 X 2 m. M X 2 Y 2 = Y 2 m. E = m. E m. M Moon

Take the square root… X is 9/10 of total distance X Y = 9 Y is 1/10 of total distance One TENTH of 3. 84 x 108 is 0. 384 x 108 Therefore, distance from the Earth where the forces are balanced is 3. 84 x 108 – 0. 384 x 108 = 3. 456 x 108 m

Syllabus Reference 11 Gravitational field of a point mass - Knowledge and use of g = Gm/r 2. - Calculation of the Earth’s mass.

Bit of Homewrok • Try the past paper question…

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 (t 2 proportional to r 3) is consistent with Newton’s law of gravitation. • Geostationary satellites.

The two equations for F = Gm 1 m 2 Newton’s Law for Gravitation r 2 F = m 2 v 2 r Centripetal force The force required to keep a planet in circular orbit 2

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

2 • Sub into equation g = Gm 1/ r 2 m 2 g g = 2 v 2 r

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

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, 000 km above the Earth • What would they be used for? • Any idea what the satellite ‘Hotbird 1’ is tracking?

Kepler’s Third Law… you gotta know this • The ratio of the period of a mass’ orbit squared (T 2) to the mean radius of it’s orbit cubed (R 3) is the same constant value k for all the planets which orbit the Sun. Planet Period (s) Average Dist. (m) T 2/R 3 (s 2/m 3) Earth 3. 156 x 107 s 1. 4957 x 1011 2. 977 x 10 -19 Mars 5. 93 x 107 s 2. 278 x 1011 2. 975 x 10 -19

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

Let’s do some substituting and rearranging… = v=2πr/T Divide each side by r 2 Multiply each side by T 2 Divide each side by Gms p v 2 = 4 π2 r 2 / T 2 Gms 4 π2 r 2 = r T 2 Gms 4 π2 = r 3 T 2 Gms 2 = 4 π r 3 T 2 4 π2 = 3 r Gms This value will be the same for all planets

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

Question 1 (Jan 2001) • The planet Mars has radius 3. 39 x 106 m and mass 6. 50 x 1023 kg. The length of a day on Mars is 8. 86 x 104 s (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

Mars has two moons, Phobos (radius of orbit 9. 38 x 103 km, 0. 319 days) and Deimos (radius of orbit 23. 5 x 103 km) c) What is the period of Deimos’ orbit? d) What are three features of a geostationary satellite?

Question 2 (Jan 2002) a) State in words, Newton’s law of Gravitation b) The period T of a satellite around a planet is related to its radius r by T 2 4 π2 = Gms r 3 where M is the mass of the planet being orbited. Show this is consistent with Newton’s Law of Gravitation

Using the data radius of Earth 6. 37 x 106 m and mass 5. 98 x 1024 kg, 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.

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

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

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 Kepler’s Law is consistent with Newton’s Law