This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 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 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....
View
Full
Document
This note was uploaded on 09/12/2009 for the course PH PH101 taught by Professor Prof.kim during the Spring '09 term at Korea Advanced Institute of Science and Technology.
 Spring '09
 Prof.Kim
 Physics, Gravity

Click to edit the document details