08_InstSolManual_PC

# 08_InstSolManual_PC - GRAVITY 8 Gm1 m2 r2 EXERCISES Section...

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GRAVITY E XERCISES Section 8.2 Universal Gravitation 12. At rest on a uniform spherical planet, a body’s weight is proportional to the surface gravity, 2 / . g GM R = Therefore, 2 ( / ) ( / )( / ) 2. p E p E E p g g M M R R = = Since / 1, / 2. p E p E M M R R = = 13. I NTERPRET In this problem we want to use astrophysical data to find the Moon’s acceleration in its circular orbit about the Earth. D EVELOP The gravitational force between two masses 1 2 and m m is given by Equation 8.1: 1 2 2 , Gm m r F = where r is their distance of separation. The acceleration of the Moon in its orbit can be computed by considering the gravitational force between the Moon and the Earth. E VALUATE Using Equation 8.1, the gravitational force between the Earth (mass ) E M and the Moon (mass ) m m is 2 E m mE GM m F r = where mE r is the distance between the Moon and the Earth. By Newton’s second law, , F ma = the acceleration of the Moon is 11 2 2 24 3 2 2 8 2 (6.67 10 N m /kg )(5.97 10 kg) 2.69 10 m/s (3.85 10 m) E m mE GM F a m r - - × × = = = = × × where we have used the astrophysical data given in Appendix E. A SSESS An alternative way to find the acceleration of the Moon as it orbits the Earth is to note that it completes a nearly circular orbit of 385,000 km radius in 27 days. Based on this information, the (centripetal) acceleration of the Moon is 2 2 2 8 3 2 2 2 4 4 (3.85 10 m) 2.73 10 m/s (27.3 86,400 s) mE mE r v a r T π π - × = = = = × × Note that since the Moon’s orbit is actually elliptical (with 5.5% eccentricity), the values based on circular orbits are reasonable. 14. If the surface gravity of the Earth were three times its present value, with no change in mass, then 2 / E GM R = 2 3 / , E E GM R where E R is the present radius. Thus, / 1/ 3 57.7% E R R = = gives the new, shrunken radius. 15. I NTERPRET In this problem we are asked to use astrophysical data to find the gravitational acceleration near the surface of (a) Mercury and (b) Saturn’s moon Titan. D EVELOP The gravitational force between two masses 1 2 and m m is given by Equation 8.1: 1 2 2 , Gm m r F = where r is their distance of separation. So for an object of mass m near a celestial body of mass M and radius R , the gravitational force between them is 2 . GMm R F = By Newton’s second law, , F ma = the gravitational acceleration near the surface of the gravitating body is 2 GM a R = 8.1 8

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8.2 Chapter 8 E VALUATE With reference to the first two columns in Appendix E, we find (a) 11 2 2 24 Mercury 2 Mercury 2 6 2 Mercury (6.67 10 N m /kg )(0.330 10 kg) 3.70 m/s (2.44 10 m) GM g R - × × = = = × (b) 11 2 2 24 2 Titan Titan 2 6 2 Titan (6.67 10 N m /kg )(0.135 10 kg) 1.35 m/s (2.58 10 m) GM g R - × × = = = × A SSESS The measured values are 2 Mercury 3.70 m/s g = and 2 Titan 1.4 m/s . g = So our results are in good agreement with the data. 16. Newton’s law of the universal gravitation (Equation 8.1), with 1 2 , m m m = = gives 1/ 2 2 6 2 11 2 2 0.25 10 N (0.14 m) 8.57 kg 6.67 10 N m /kg Fr m G - - × = = = × (Newton’s law holds as stated, for two uniform spherical bodies, if the distance is taken between their centers.) 17. I NTERPRET In this problem we want to find the gravitational force between the astronaut and the space shuttle.
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