Gravitation: Newton's Law, Kepler's Laws (PDF)

GRAVITATION If two bodies of masses M and m whose centres are separated by a distance r, interact M...

GRAVITATION If two bodies of masses M and m whose centres are separated by a distance r, interact M m with each other, then there exists a force of r interaction between them, which makes them to either attract or repel one another. This force acts along their common centres. a) Newton’s law of gravitation: Gives the amount of force which exists between the interacting bodies. 𝑀𝑚 It is given by 𝐹= 𝐺 ; Unit: N 𝑟2 1 This implies that 𝐹 ∝ ; it is an inverse-square law 𝑟2 where F – the force established between the two interacting masses (in N); r – distance between the centres of the interacting masses (in m); G – universal gravitational constant (G = 6.67 × 10-11 Nm2kg-2) 𝑀 b) Gravitational Potential, V: 𝑉 = −𝐺 or V = -gr Unit: Nmkg-1 𝑟 where M – the mass of the planet (in kg); r – the radius of the planet (in m); and g – acceleration due to gravity on the planet (in ms -2) c) Mass M of a Planet 𝑅2 The mass M of a planet can be calculated using 𝑀=𝑔 𝐺 where all the parameters maintain their meanings 2𝐺𝑀 d) Escape velocity, v: 𝑣= √ ; or 𝑣 = √2𝑔𝑟 ; or 𝑣 = √2|𝑉| 𝑟 where |V| - magnitude of V, the gravitational potential (in Nmkg -1) M – mass of planet (in kg); r – radius of planet (in m) Kepler’s Laws of Planetary Motion a) Law of Orbits o Each planet moves in an elliptical orbit with the Sun at one focus of the ellipse b) Law of Areas o A line joining the Sun to a given planet sweeps out equal areas in equal times  c) Law of Periods o The squares of the periods of revolution of the planets are proportional to the cubes of their mean distances from the Sun PRACTICE EXERCISE 1. The centre of a planet of mass 3.3 × 1023 kg is at a distance of 58.0 × 109 m from the centre of the Sun. If the mass of the sun is 2.0 × 1030 kg, what is the force binding the planet to the sun? Take G = 6.67 × 10-11 Nm2kg-2 Answer: F = 1.31 × 1022 N Solution: Data: m = 3.3 × 1023 kg, r = 58.0 × 109 m, M = 2.0 × 1030 kg, F=? 𝑀𝑚 2.0 ×1020 × 3.3 ×1023 Applying 𝐹= 𝐺 , implies that 𝐹 = 6.67 × 10−11 × 𝑟2 (58.0 × 109 )2 Hence F = 1.31 × 1022 N 2. Calculate the escape velocity for a planet of radius 6.4 × 10 6 m and acceleration due to gravity 9.8 ms-2. Answer: v = 11200 ms-1 or v = 11.2 kms-1 Solution: Data: r = 6.4 × 106 m, g = 9.8 ms-2 v=? Applying 𝑣 = √2𝑔𝑟; implies that 𝑣 = √2 × 9.8 × 6.4 × 106 Hence v = 11200 ms-1 3. If the acceleration due to gravity is 9.8 ms-2 and the radius of the earth is 6400 km, calculate the value for the mass of the earth. Take G = 6.67 × 10-11 Nm2kg-2 Solution: Data: R = 6.40 × 106 m, M =? 𝑀 𝑅2 9.8 ×( 6.40 × 106 )2 Applying 𝑔 = 𝐺 𝑅2 , 𝑀=𝑔 implies that 𝑀= 𝐺 6.67 ×10−11 Hence M = 5.99 × 1024 kg 4. Two lead balls whose masses are 5.20 kg and 0.25 kg are placed with their centres 50.0 cm apart with what force do they attract each other? Solution: Data: M = 5.20 kg, m = 0.25 kg, r = 0.50 m, F=?  𝑀𝑚 5.2 × 0.25 Applying 𝐹= 𝐺 , implies that 𝐹 = 6.67 × 10−11 × 𝑟2 (0.5 )2 Hence F = 3.46 × 10-10 N