Human gravity Newton’s law of universal gravitation Mass, weight and acceleration due to gravity Check-point Newton’s law of universal gravitation Book 2 Section 8.1 Newton's law of universal gravitation

P.2 Book 2 Section 8.1 Newton's law of universal gravitation If every object is attracted to each other by gravitational force, Human gravity why do we not ‘stick’ together with each other? The gravitational force between us is too small!

P.3 Book 2 Section 8.1 Newton's law of universal gravitation 1 Newton’s law of universal gravitation Every particle attracts every other particle with a gravitational force. along the line joining the 2 particles forming an action-and-reaction pair: F –F–F equal in magnitude but opposite in direction

P.4 Book 2 Section 8.1 Newton's law of universal gravitation 1 Newton’s law of universal gravitation Gravitational force F between 2 particles: G : universal gravitational constant, 6.67  10 –11 N m 2 kg –2 Newton’s law of universal gravitation F =F = Gm1m2Gm1m2 r 2

P.5 Book 2 Section 8.1 Newton's law of universal gravitation 1 Newton’s law of universal gravitation can be applied to spherically symmetric objects assume the total mass of each object is concentrated at its centre Example 1 Gravitational force Newton’s law of universal gravitation: r = distance between their centres

P.6 Book 2 Section 8.1 Newton's law of universal gravitation Example 1 Gravitational force Assume Mary and John are two spherically symmetric objects. G = 6.67  10 –11 N m 2 kg –2 Given: Mass of the Earth M E = 5.98  kg Mass of the Sun M S = 1.99  kg Radius of the Earth R E = 6370 km

P.7 Book 2 Section 8.1 Newton's law of universal gravitation Example 1 Gravitational force (a)Mass of Mary = 50 kg Mass of John = 70 kg Distance between them = 2 m Gravitational force between them = ? F = 6.67  10 –11  50  702 = = 5.84  10 –8 N Gm1m2Gm1m2 r 2

P.8 Book 2 Section 8.1 Newton's law of universal gravitation Example 1 Gravitational force (b) Mary is on the Earth surface. Find the gravitational force acting on her by the Earth  10 –11  50  5.98  (6.37  10 6 ) 2 = = 491 N (to the centre of the Earth) F =F = Gm1MEGm1ME RE 2RE 2

P.9 Book 2 Section 8.1 Newton's law of universal gravitation Example 1 Gravitational force (ii) Find the gravitational force acting on her by the Sun (1.5  m from the Earth)  10 –11  50  1.99  (1.50  ) 2 = = N (to the centre of the Sun) F = Gm1MSGm1MS r 2

P.10 Book 2 Section 8.1 Newton's law of universal gravitation 2 Mass, weight and acceleration due to gravity The weight is actually the gravitational force acting on the object by the Earth. From the law of universal gravitation, (1) M E : mass of Earth m : mass of object W = GMEmGMEm r 2

P.11 Book 2 Section 8.1 Newton's law of universal gravitation From Newton’s second law of motion, W = mg (2) Substitute (2) into (1), 2 Mass, weight and acceleration due to gravity (3) mg = GMEmGMEm r 2r 2 g = GMEGME r 2r 2

P.12 Book 2 Section 8.1 Newton's law of universal gravitation 2 Mass, weight and acceleration due to gravity If an object is close to the Earth surface, we can use (3) to find the theoretical value of g. Take M E = 5.98  kg, r = 6370 km  10 –11  5.98  (6.37  10 6 ) 2 = = 9.83 m s –2 Example 2 Weight of Chang’e 1 g =g = GMEGME r 2r 2

P.13 Book 2 Section 8.1 Newton's law of universal gravitation Example 2 Weight of Chang’e 1 G = 6.67  10 –11 N m 2 kg –2 Mass of the Earth M E = 5.98  kg Radius of the Earth R E = 6370 km The first lunar probe of China, Chang’e 1, has a mass of 2350 kg. Given:

P.14 Book 2 Section 8.1 Newton's law of universal gravitation Example 2 Weight of Chang’e 1 (a) Find the weight of Chang’e 1 when it is at rest on launching platform. On the platform, r = R E  10 –11  5.98   2350 (6.37  10 6 ) 2 = = 2.31  10 4 N W =W = GMEmGMEm r 2r 2

P.15 Book 2 Section 8.1 Newton's law of universal gravitation Example 2 Weight of Chang’e 1 (b) Find the weight of Chang’e 1 when it is 2R E above the surface of the Earth. r = R E + 2R E = 3R E 6.67  10 –11  5.98   2350 (3  6.37  10 6 ) 2 = = 2.57  10 3 N W =W = GMEmGMEm r 2r 2

P.16 Book 2 Section 8.1 Newton's law of universal gravitation Example 2 Weight of Chang’e 1 (b) Alternative Method: By proportion, W =  answer in (a) = 2.57  10 3 N =  2.31  10 4

P.17 Book 2 Section 8.1 Newton's law of universal gravitation Check-point 1 – Q1 Find F acting on the Earth by the Sun. Mass of the Earth = 5.98  kg The Earth and the Sun are 1.50  m apart.) F = Gm1m2r 2Gm1m2r 2 (Mass of the Sun = 1.99  kg = 6.67  10 –11  1.99   5.98  (1.50  ) 2 = 3.53  N

P.18 Book 2 Section 8.1 Newton's law of universal gravitation Check-point 1 – Q2 The gravitational force acting on the Earth by a particle is much smaller than that acting on the particle by the Earth. (T / F) True or False:

P.19 Book 2 Section 8.1 Newton's law of universal gravitation Check-point 1 – Q3 Chang’e 1 orbits round the Moon 200 km above its surface. If mass of Chang’e 1 = 2350 kg, weight = ? Given: Mass of Moon = 7.35  kg Radius of Moon = 1.74  10 6 m W =W = GMmr 2GMmr  10 –11  7.35   2350 (1.74   10 3 ) 2 = = 3060 N

P.20 Book 2 Section 8.1 Newton's law of universal gravitation As the weight of an object on the Moon is that on the Earth, a vehicle needs less driving force on the Moon Check-point 1 – Q4 (T / F) True or False:

P.21 Book 2 Section 8.1 Newton's law of universal gravitation The End