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a) SO, given all the above, Derive Fgr = f(r) for which Fgr Sstands for the net gravitational force (dependent variable) exerted on some particle of mass, m, by all the bits of a large SOLID sphere of mass M, as determined byr, displacement from the center of a solid sphere (independent variable) -- assuming thatr =

a) SO, given all the above, Derive Fgr = f(r) for which Fgr Sstands for the net gravitational force (dependent variable) exerted on some particle of mass, m, by all the bits of a large SOLID sphere of mass M, as determined byr, displacement from the center of a solid sphere (independent variable) -- assuming thatr =

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IV. Subterranean Harmonic Tho statomonts immodiatoly to follow (ovon whon long and cormplicatod) aro considorod (in this contoxt) GIVEN. You may assumo and roly on thom for tho problom/proof to follow a bit furthor down. Nota: In somo casos. 'GIVEN' might moan 'solf-ovidont' or 'obvious', but in othor casos, it might not. GIVEN might not moan 'obvious': it can simply moan 'somohow ostablishod prior to this discussion'. GIVEN (for this contoxt) > 1) Planot Earth can bo troatad as ono largo SOLID SPHERE of UNIFORM (constant) DENSITY. with constant ('known') mass M. constant ('known') radius R. 2) Given ANY TWO POINT MASSES, say m, and m2. then m, will necessarily oxert a gravitational force onto m2and thereby pull m2 directly toward m1. (*If we wish to be caroful and precise in koeping track of directions, we can do the following: Lot r be a displacomont voctor that points directly from m, to m2 and is precisely as long as the distance botween m1 and m2. then the customary scalar quantity r. reforring to distance between massas. can be undorstood as the pure magnitude of the voctor r. while the possibly loss familiar r is understood as the pure direction of r) Henco, the gravitational pull exerted by m, onto m2 is given by force given by: F=- Gm,m, - 2 This Claim is symmetric (applios idontically to the pull exortod by m2 onto m1). but is always and only a statement about the interaction which arisos BETWEEN 2 (infinitosimally small) POINTS OF MASS ('particlas'). Page 7 of 10 а) SO, given all the above, Derive Fgr = f(r) for which Fgr stands for the net gravitational force (dependent variable) exerted on some particle of mass, m, by all the bits of a large SOLID sphere of mass M, as determined byr, displacement from the center of a solid sphere (independent variable) -- assuming thatr =<R, that is: CS Scanned with CamScanner that the particle is located somewhere within the solid sphere.
Transcribed Image Text:IV. Subterranean Harmonic Tho statomonts immodiatoly to follow (ovon whon long and cormplicatod) aro considorod (in this contoxt) GIVEN. You may assumo and roly on thom for tho problom/proof to follow a bit furthor down. Nota: In somo casos. 'GIVEN' might moan 'solf-ovidont' or 'obvious', but in othor casos, it might not. GIVEN might not moan 'obvious': it can simply moan 'somohow ostablishod prior to this discussion'. GIVEN (for this contoxt) > 1) Planot Earth can bo troatad as ono largo SOLID SPHERE of UNIFORM (constant) DENSITY. with constant ('known') mass M. constant ('known') radius R. 2) Given ANY TWO POINT MASSES, say m, and m2. then m, will necessarily oxert a gravitational force onto m2and thereby pull m2 directly toward m1. (*If we wish to be caroful and precise in koeping track of directions, we can do the following: Lot r be a displacomont voctor that points directly from m, to m2 and is precisely as long as the distance botween m1 and m2. then the customary scalar quantity r. reforring to distance between massas. can be undorstood as the pure magnitude of the voctor r. while the possibly loss familiar r is understood as the pure direction of r) Henco, the gravitational pull exerted by m, onto m2 is given by force given by: F=- Gm,m, - 2 This Claim is symmetric (applios idontically to the pull exortod by m2 onto m1). but is always and only a statement about the interaction which arisos BETWEEN 2 (infinitosimally small) POINTS OF MASS ('particlas'). Page 7 of 10 а) SO, given all the above, Derive Fgr = f(r) for which Fgr stands for the net gravitational force (dependent variable) exerted on some particle of mass, m, by all the bits of a large SOLID sphere of mass M, as determined byr, displacement from the center of a solid sphere (independent variable) -- assuming thatr =<R, that is: CS Scanned with CamScanner that the particle is located somewhere within the solid sphere.
b. Using the rosult you derived in (a). above. imagine that a tunnel is carved out from one place on the Earth's surface to another place on Earth's surface. Imagine that this tunnel passes through the Earth's center so that it is a diameter. A small mass m, such as a boulder or a subway car, is dropped into one end of this tunnel. The tunnel is empty of anything frictional-including air. Assuming that m free-falls through this gravitational tunnel. the koy question becomes: How much time will it take for the mass to reach the other side? Hint #1: This key question, above, is where all your work for (a) becomes worth it. This is where your understanding of simple harmonic motion becomes relevant. This is where you see why simple harmonic motion is such a sweet concept. i. First, find this time as a general exprossion: as a function of the given and fundamental constants (G. M. R). ii. Hint #2: Evaluate your function in order to get an actual numerical measurement· - an actual number of MINUTES, in this case, as an answer for time across. In order to obtain a numerical answer for time, use the standard numerical values for Earth's characteristics: MẸ~ 6.0 × 10"kg E RE 6.4× 10 meters. с. Given all that you have derived above (particularly in part (a)), and given all you solved in Part II of this exam, Determine: How far from the center of the Earth is the pendulum of Part II ('Hooke on Hook')? That is, at what fraction of Rp do we find the pendulum in Part II? Hint: compute g, according to Fattly Belge g to r, according to Part III.
Transcribed Image Text:b. Using the rosult you derived in (a). above. imagine that a tunnel is carved out from one place on the Earth's surface to another place on Earth's surface. Imagine that this tunnel passes through the Earth's center so that it is a diameter. A small mass m, such as a boulder or a subway car, is dropped into one end of this tunnel. The tunnel is empty of anything frictional-including air. Assuming that m free-falls through this gravitational tunnel. the koy question becomes: How much time will it take for the mass to reach the other side? Hint #1: This key question, above, is where all your work for (a) becomes worth it. This is where your understanding of simple harmonic motion becomes relevant. This is where you see why simple harmonic motion is such a sweet concept. i. First, find this time as a general exprossion: as a function of the given and fundamental constants (G. M. R). ii. Hint #2: Evaluate your function in order to get an actual numerical measurement· - an actual number of MINUTES, in this case, as an answer for time across. In order to obtain a numerical answer for time, use the standard numerical values for Earth's characteristics: MẸ~ 6.0 × 10"kg E RE 6.4× 10 meters. с. Given all that you have derived above (particularly in part (a)), and given all you solved in Part II of this exam, Determine: How far from the center of the Earth is the pendulum of Part II ('Hooke on Hook')? That is, at what fraction of Rp do we find the pendulum in Part II? Hint: compute g, according to Fattly Belge g to r, according to Part III.
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