HCI 07 Gravitation Solutions (Lecture Examples)

Example 1: Finding Resultant Gravitational Force Three identical masses, each of mass m, are located on a table at the corners of an equilateral triangle of side d. Determine the resultant gravitational force on mass C due to masses A and B. Solution: Magnitude of force on C due to B 2 B on C 2 GmF d Also, force on C due to A = force on C due to B Hence by vector addition, the resultant force = 2 FB on C cos 30o 2 2 3Gm d directed horizontally towards the left. A B C y x Concept: If several particles are present, each pair will experience a mutual gravitational attraction. The resultant gravitational force on a given particle is the vector sum of the separate attractive forces acting on the particle due to individual particles interacting with it. FB on C FA on C

2 Example 2: Gravitational Field Strength near the Surface of the Earth Assume that the Earth is not rotating and is a perfect sphere of uniform density. (i) Calculate the gravitational field strength of the Earth a t its surface. (ii) Given that Mount Everest is 8.85 km high, compute the grav itational field strength of Earth at the peak of Mount Everest. (iii) Comment on your values computed above. You are given the following information:  Mass of the Earth, mE = 5.98 x 1024 kg  Average radius of Earth, rE = 6.37 x 106 m Solution: (i) The gravitational field strength at the surface of the Eart h, gsurface = (ii) The gravitational field strength at the peak of Mount Ever est, geverest = (iii) Comparing the values compu ted in (i) and (ii), we see that  gsurface > geverest by only approx. 0.03 m s-2 (about 0.3%).  Hence, for the purpose of calculations, g may be taken to be constant near the Earth surface. Furthermore, for heights h near the surface of the Earth such that h << rE, the value of g may be taken to be constant. rE h = 8.85 km 2 E E r mG = (6.67 x 10-11 N m2 kg-2) 24 62 5.98 10 kg (6.37 10 m)     = 9.83 m s-2. 2() E E mG rh = (6.67 x 10-11 N m2 kg-2) 24 63 2 5.98 10 kg (6.37 10 m 8.85 10 m)      = 9.80 m s-2

3 Example 3: How Earth’s rotatio n affects the weight measured Consider a crate of mass m resting on a weighing machine at the equator. Take the Earth to be a uniform sphere of mass M and radius R, spinning on its axis. (a) On the diagram, draw all the forces acting on the crate. (b) Write down an expression fo r the reading of the weighing machine supporting this mass. (c) Write down an expression for the ‘apparent’ gravitational acceleration indicated by the weighing machine. Solution: (b) By Newton’s second law (considering the crate): c gc gc Fm a FN m a NF m a     By Newton’s 3rd law, the force by crate on scale = N. (c) Apparent gravitational field strengt