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 strength ' gc c c Fm a mg maNgg amm m      What is the value of the centripetal acceleration? 2 26 2 26.4 10 0.034 m s24 3600 car       Hence the apparent gravitational field strength is 9.81 – 0.034 = 9.78 m s-2 normal contact force, N [this is force by scale on crate] gravitational force Fg

4 Example 4 Investigating satellite motion Consider a satellite of mass m revolving around a larger mass M in a circle of radius r. Determine the relationship between the radius of the orbit and (a) the velocity of the satellite; (b) the period of rotation. Solution: As the satellite revolves, its velocity keeps changing as the direction of motion keeps changing. Hence, the satellite is accelerating. From the topic of Circular Motion, we know that a resultant force must be providing this centripetal acceleration. The gravitational attraction of the large mass M on the satellite keeps the satellite in orbit, i.e., the gravitational force acting on the satellite (by the mass M) provides for the centripetal force required for the satellite to maintain circular orbit about the mass M. (a) By Newton’s 2nd Law, 2 2 1 cFm a GMm v m rr GMv r v r      (b) By Newton’s 2nd Law, 2 2 2 32 23 4 cFm a GMm mrr GM rT Tr        Conclusion: For an orbit of radius r, there is only one allowed value of T and v. M m r

5 Example 5: How do you ‘weigh’ the Earth? When Newton first discovered his Law of Gravitation, he had neither the value of G nor the mass of the Sun. In 1798, a hundred and twenty-one years after Newton proposed t he Universal Law of Gravitation, Henry Cavendish conducted the first experiment to measure the force o f gravity between masses in the laboratory, and obtained accurate values for the gravitational constant, G, and the mass density of the Earth. With the value of G, it was possible to obtain a value for the mass of the Earth. By considering the motion of the moon about the Earth, show tha t the mass of the Earth me is approximately 6 x 1024 kg. You may assume that distance between Earth and the moon r = 4.0 × 108 m. Solution: The gravitational force acting on the moon by the Earth’s gravitational field provides for the centripetal force required by the moon to move in orbit about the Earth. By Newton’s 2nd Law, Substituting for G, rm and since we know that the moon takes approximately about 1 month to make one complete revolution around the Earth, Hence, we have found the mass of the Earth! The Cavendish Experiment - Sixty Symbols me mm r Fg Earth moon Fg = mm ac  2 emmmG r = mm r m2 = 2 2   m m mr T  me = 23 2 4 m r GT   me = 23 2 4 m r GT  = 28 3 2 4( 4 1 0 m ) ( 3 02 46 06 0 s )G    = 5.64 x 1024 kg ~ 6 x 1024 kg

6 Example 6: Geostationary Satellite A communications satellite of mass m is placed in a circular geostationary orbit. Find its height above the Earth’s surface. (Take radius of the Earth to be 6.38 x 106 m and the mass of the Earth to be 5.98 x 1024 kg). Solution: The gravitational force on the satellite provides the required centripetal force. (NOTE: the statement above is required to explain the following steps below. Please write this in exams!) By Newton’s second law:  2 2 2 32 211 242 3 22 7 767 4 6.67 10 5.98 10 24 3600 44 4.23 10 m 4.23 10 6.38 10 3.59 10 m cFm a GMm mrr GM rT GMTr r h                  

7 Example 7: Gravitational Potential Energy of a System of 3 Mass es A system consists of three particles, each of mass m located at the corners of an equilateral triangle of side x. Determine the gravitational potential energy of the system. Solution: In order to assemble the three particles at the corners of the equilateral triangle, we bring each particle (from infinity) to its assigned position one after another. The work done by us while bringing each particle to its position is equal to the change in the gravitational potential energy of the system. Hence the total change in the gravitational potential energy of the system U is the net work done W net to position all three particles. ∆𝑈 ൌ ∆𝑊௡௘௧ Since the gravitational potential energy in the system is zero initially when all the particles are infinitely far away, 𝑈௙ െ0ൌ𝑊 ଵ ൅𝑊ଶ ൅𝑊ଷ Note that when positioning particle 3, both particle 1 and 2 are already positioned and they exert gravitational forces on the third particle. 13 2312 12 13 23 0 () () ()f Gm m Gm mGm mU rr r           Since all particles have the same mass and they are at the vertices of an equilateral triangle of side x, 𝑈௙ ൌെ 3𝐺𝑚ଶ 𝑥 ଷீ௠మ ௫ would be the work needed to separate the particles by an infinite distance. Example 8: Gravita