1. Newton 1642-1727 Anything with mass attracts anything else with mass. The size of that attraction is given by my Law of Gravitation: Fg = Gm 1 m 2 r 2 …where m 1 and m 2 are the masses of the two objects (in kg), r is the distance between their centers (in m) and G is “The Universal Gravitational Constant” (= 6.67 x 10 -11 Nm 2 kg -2 ). Fg  m 1 m 2 r 2 OR *This is a field force, not a contact force.

2.  Determined experimentally by Henry Cavendish in 1798  Two identical masses were suspended and free to rotate  Two identical larger masses were brought near them, force of attraction measured The light beam and mirror serve to amplify the motion

5. 1) Two objects each with a mass of m are separated by a distance d. The force of gravity between them is 100 N. What would the force of gravity be on these objects if… a) 1 of the masses doubled b) Distance between the masses quadrupled c) Distance between the masses was halved F  m 1 m 2 r 2  (2) (1) 1 2 = 2x’s more F  m 1 m 2 r 2  (1) (1) 4 2 = 1/16 as much F  m 1 m 2 r 2  (1) (1) (½) 2 = 4 x’s more

6. RERE M MEME Mass of Sun RSRS 1.99 x 10 30 kg MSMS

7. 2) What is the force of attraction between a 70kg mass and an 80kg mass if they are 150cm apart? Fg = Gm 1 m 2 r 2 = (6.67x10 -11 N m 2 /kg 2 ) (70kg) (80kg) (1.5 m) 2 = 1.66 x 10 -7 N = 3.74 x10 -11 2.25

8. 3) a) Calculate the mass of the Earth given that a mass of 1kg has a weight of 9.8N on the Earth (radius 6370km). b) Use this value to calculate the gravitational attraction between the Earth and the moon. 9.8N= (6.67x10 -11 N m 2 /kg 2 ) (1 kg) (M E ) (6.37x10 6 m) 2 M E = 5.96 x 10 24 kg F= 1.98 x 10 20 N = 6.67x10 -11 ) ( 5.96x10 24 ) (7.35x10 22 ) (3.84x10 8 m) 2

9. The Earth attracts the man and the man attracts the Earth – a Newton III pair of forces where both are gravitational.

10. Force per unit mass experienced by a small test mass placed in that field Vector quantity: has direction (towards the center of a mass) Gravitational field strength and acceleration due to gravity are the same thing. Field lines show the direction a test mass would move as a result of the force applied on it.

11. A uniform gravitational field is one where the field lines are always the same distance apart - this is almost exactly true close to the Earth's surface (Figure 1(a)). However if we move back and look at the planet from a distance the field lines clearly radiate outwards (Figure 1(b)), getting further apart as the distance from the Earth increases. When viewed from an even greater distance the complete field can be seen (Figure 1(c)). Such a field is called a radial field - the field intensity (g) decreasing with distance.

12. Consider a man on the Earth: Man’s weight = mg BUT we know that this (weight) is equal to his gravitational attraction, so… GMm = mg r 2 GM = g r 2 Therefore:

13.  Acceleration due to gravity  g will vary with altitude  Mass of Earth: 5.98 x 10 24 kg  Average Radius: 6.37 x 10 6 m

14. Gravitational Potential V Work or done (or energy expended) per unit mass in moving a small mass from infinity to that point Since E p = mg  h, gravitational potential is equal to Equipotential lines NOT ON REFERENCE TABLE! OR Lines of equal potential Perpendicular to Field lines scalar quantity

15. Gravitational Potential V In general ΔV = potential difference(J/kg) potential gradient= field strength For a radial field V = gravitational potential at a distance r from mass m Since E p = mg  h mg  h =  V * m g =  V /  h g =  V /  r Since g = G m / r 2 and g = -  V /  r -  V /  r = G m / r 2 V = -G m / r Negative sign indicates work done against the gravitational field

16. Calculate the gravitational field strength at points A & B 200 kg 800 kg A B 3m 7m4m For A g = Gm/r 2 = 6.67x10 -11 * 200kg / (3m) 2 = 1.48 x 10 -9 N kg -1 g = Gm/r 2 = 6.67x10 -11 * 800kg / (14m) 2 = 2.72 x 10 -10 N kg -1 g = 1.48 x 10 -9 N kg -1 + 2.72 x 10 -10 N kg -1 = 1.75 x 10 -9 N kg -1 For B g = Gm/r 2 = 6.67x10 -11 * 200kg / (7m) 2 = 2.72 x 10 -10 N kg -1 g = Gm/r 2 = 6.67x10 -11 * 800kg / (4m) 2 = 3.34 x 10 -9 N kg -1 g = 3.34 x 10 -9 N kg -1 - 2.72 x 10 -10 N kg -1 = 3.06 x 10 -9 N kg -1 right

17. The radius of Mars is 3.4 x10 6 m and the magnitude of the gravitational field strength at a height of 1.2 x 10 6 m above its surface is 1.8 N kg –1. What is the magnitude of the gravitational potential at this height? g = - ΔV / Δ r Δ r = 3.4 x10 6 m + 1.2 x 10 6 m = 4.6 x 10 6 m 1.8 N kg –1 = - V / 4.6 x 10 6 m V = - 8.3 x 10 6 J kg -1

18.  PE = mgh is valid only near the earth’s surface  For objects high above the earth’s surface, an alternate expression is needed  Zero reference level is infinitely far from the earth’s center Since ΔE p = m ΔV g and V g = - GM / r E p = -GMm/ r

20.  The escape speed is the speed needed for an object to soar off into space and not return  When the object reaches this point, it would have a total Energy of zero  Since E T = KE + PE E T = ½ mv 2 – GMm/r ½ mv 2 = GMm/r ½ v 2 = GM/r v 2 = 2GM/r

21.  For the earth, v esc is about 11.2 km/s  Note, v is independent of the mass of the object Mass of Earth (or other astonomical body Radius of Earth (or other astonomical body

22.  All planets move in elliptical orbits with the Sun at one focus (2 nd focus is empty)  A line drawn from the Sun to any planet will sweep out equal areas in equal times  Area from A to B and C to D are the same

23.  The square of the orbital period of any planet is proportional to cube of the average distance from the Sun to the planet.  For an orbit around the Sun, K S = 2.97x10 -19 s 2 /m 3  K is independent of the mass of the orbiting body

24.  Mass of the Sun or other celestial body that has something orbiting it  Gravity provides the centripetal force for circular motion  Assuming a circular orbit is a good approximation

25. Fc = Fg, so mv 2 / r = GMm/ r 2, where M is the mass of the sun and m is the mass of a planet. v 2 = GM/ r In a circle, v = 2  r / T, so (2  r/T ) 2 = GM/ r 4   r 2 /T 2 = GM/ r GM T 2 = 4   r 3 T 2 / r 3 = 4   / GM Since 4   / GM is constant, T 2 / r 3 = constant, so T 1 2 / r 1 3 = T 2 2 / r 2 3

26.  Since F c = F g mv 2 /r = GMm/ r 2  mv 2 = GMm/ r  Multiplying each side by ½ :  ½ mv 2 = ½ ( GMm / r) = GMm/ 2r  Therefore E K = GMm/2r  Since E K = ½ mv 2  v 2 = GM / r  v orbit = √GM/r

27.  E T = E K + E P = GMm/2r - GMm/ r = GMm/2r - 2GMm/2 r = - GMm/2r