1. Gravitational Fields Newton’s Universal Law F  Mm & F  1/r 2 What are the units of G, the Universal Gravitational Constant? G = 6.672 x 10 -11 Nm 2 kg -2

3. G & g On the surface of the Earth, F = mg and F = GmM e /r 2 so g = Gm e /r 2 You might sometimes see a negative sign in this equation. That is because r is measured out from the centre of the Earth whereas g is directed towards the centre It is an interesting fact that here we have equated inertial and gravitational masses. Why? It would seem that mass is intrinsically linked to the way objects move through space.

4. g / ms -2 r / m rere Conjecture - as the core is more dense than the rest. Above the Earth. g  1/r 2 Inside the Earth if the density is constant.. What does conjecture mean?

5. You have raised its potential energy to zero. We call all the potential energy in the field negative and say that it rises to zero at infinity. (r 2 >r 1 )

7. Example 1 Calculate the mass of the moon if its radius is 1.76x10 6 m and the gravitational field strength at its surface is 1.7Nkg -1 (G=6.67x10 -11 Nm 2 kg -2 ). Solution 1 g = -GM/r 2 -1.7 = -6.67x10 -11 x M / (1.76x10 6 ) 2 M = 1.7 x (1.76x10 6 ) 2 / 6.67x10 -11 M = 7.89 x 10 22 kg (Actual value is 7.3 x 10 22 kg. The radius is not uniform - an interesting concept to consider when predicting the exact time and duration of an eclipse.)

8. Example 2 Calculate the gravitational potential on the surface of Mercury due to: a) Mercury(M m = 3x10 23 kg) b) The Sun(M s = 2x10 30 kg) c) The Earth(M e = 6x10 24 kg) (R m = 2.4x10 6 m, Mercury to Sun = 5.8x10 10 m, Earth to Mercury =9.2x10 10 m) Solution 2 We are looking for the potential V which is given generally by V = GM / r a) b) c)

9. Example 3 Calculate the gravitational potential difference between a point at sea level and the summit of Mount Everest which is 8.9km above sea level. Solution 3 = 8.7x10 4 J Nb GM = 6.67x10 -11 x 6x10 24 = 4x10 14 1/r 1 - 1/r 2 = -1/(6.4x10 6 + 8.9x10 3 ) - -1/6.4x10 6 = 2.17x10 -10  V = 4x10 14 x 2.17x10 -10

10. 0 -2.45 -54.3 -61.8 -62.7 -62.8  10 4 10 3 10 2 10 1 0 Potential / MJkg -1 Height / km Example 4 a)How much potential energy do we need to raise this 2kg mass from 0 to 10km ? b)......from 0 to  ? c)......from 10 2 to 10 4 km? d)......FROM 10 3 TO 10 1 km? Solution 4 V = E p / m so  E p = m  V a)  E p = 2x( -62.7--62.8) = 2x0.1 = 0.2MJ b)  E p = 2x( 0--62.8) = 2x62.8 = 125.6MJ c)  E p = 2x( -2.45--61.8) = 2x59.3= 118.7MJ d)  E p = 2x( -62.7--54.3) = 2x8.4 = -16.8MJ

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