Question
help with practice problem.answer is provided. don't know if I am approaching the problem wrong
tud Known Unknown (a) Ma SOLUTION Part (a) of Eart h, r = 1.50 1011 m; orbital radius of Mercury, r 579 x 1010 m. e Sun, M?()Period of Mercury, T? 123 KEPLERS LANS oF OREt on 12-7 for the mass of the Sun: Woa 2. calculate the period of the Earth in seco GT 365.24 days 24h (3600 s Substitute numerical values in the for the mass of the Sun obtained in stPression e expression -316 10s tained in Step 1 day GT2 (6.67 x 101,N-m2kgz )(3.16 × 10'); 2.00 x 10"'kg 9m2(1.50 x 10" m)) Part (b) 4. Substitute r = 5.79 × 1010 m into Equation 12.7. In addition, use the m in part (a): ass of the Sun obtained 2. V(6.67 × 10-11N-kg,1200 10" kg) 7.58 x 10°,-0.240s . 87.7 days INSIGHT In part (a), we find that the mass of the Sun is almost a million times greater than the massof the Earth In tact, accounts for 9 a shorter year than the Earth. 9.9% of all the mass in the solar system. In part(b) we see that Mercury with itss aller titah PRACTICE PROBLEM Venus orbits the Sun with a period of 1.94 x 10's. What is its average distancefom the Sun? Answer : r= 1.08 × 1011 m] ome related homework problems: Problem 25, Problem 28 ina System (GPS) are in relatively low or- rmination of an Constant distance fron
Explanation / Answer
T = (2*pi/sqrt(GM))r^1.5
G = 6.67 x 10^-11
Ms = 2 x 10^30
T = 1.94 x 10^7
r = 8.4 x 10^10 m
