chap 8 - GRAVITY 8 Gm1m2 r2 EXERCISES Section 8.2 Universal...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
8.1 GRAVITY EXERCISES Section 8.2 Universal Gravitation 12. At rest on a uniform spherical planet, a body’s weight is proportional to the surface gravity, gG M R = /. 2 Therefore, (/)( / ) ( /) . gg MM RR pE p E E p == 2 2 Since MM R R p E /, / . 12 13. INTERPRET In this problem we want to use astrophysical data to find the Moon’s acceleration in its circular orbit about the Earth. DEVELOP The gravitational force between two masses mm and is given by Equation 8.1: F Gm m r = 2 , 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. EVALUATE Using Equation 8.1, the gravitational force between the Earth (mass M E ) and the Moon (mass m m ) is F GM m r Em mE = 2 where r mE is the distance between the Moon and the Earth. By Newton’s second law, Fm a = , the acceleration of the Moon is a F m GM r m E mE = ×× 2 11 2 2 6 67 10 5 97 10 (. / ) Nmk g 24 82 32 385 10 269 10 kg m ms ) ) ./ × where we have used the astrophysical data given in Appendix E. ASSESS 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 a v r r T mE mE = × × 2 2 2 28 4 4 3 85 10 27 3 86 400 π ) , m s m/s ) . 2 273 10 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 GM R E / 2 = 3 2 GM R EE where R E is the present radius. Thus, RR E // . % 1 3 57 7 gives the new, shrunken radius. 15. INTERPRET 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. DEVELOP The gravitational force between two masses and is given by Equation 8.1: F Gm m r = 2 , 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 F GMm R = 2 . By Newton’s second law, a = , the gravitational acceleration near the surface of the gravitating body is a GM R = 2 8
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
8.2 Chapter 8 EVALUATE With reference to the first two columns in Appendix E, we find (a) g GM R Mercury Mercury Mercury 2 Nm == × (. 667 10 11 2 2 2 4 62 0 330 10 244 10 370 /) ( . ) ) ./ kg kg m m × × = s 2 (b) g GM R Titan Titan Titan 2 Nmk g × / ) 11 2 2 ) ) 0 135 10 258 10 135 24 2 × × = kg m ms ASSESS The measured values are g Mercury = 2 and g Titan = 14 2 . So our results are in good agreement with the data. 16. Newton’s law of the universal gravitation (Equation 8.1), with mmm 12 , gives m Fr G × ×⋅ 26 2 11 2 025 10 014 .( . ) . / . / kg kg 2 857 = (Newton’s law holds as stated, for two uniform spherical bodies, if the distance is taken between their centers.) 17. INTERPRET In this problem we want to find the gravitational force between the astronaut and the space shuttle.
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 15

chap 8 - GRAVITY 8 Gm1m2 r2 EXERCISES Section 8.2 Universal...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online