Question

Now M is the sum of the two masses in units of the solar mass .e. the mass of our Sun), while a is still in AU and P in years. An important application of Newton's generalization of Kepler's third law is being able to dete mine mass of a central body based on the motion of a satellite around that body. If the satellite is much less massive than the body it's orbiting, then M is essentially equal to just the mass of the central body (this makes sense think about the masses of the planets vs. the Sun's mass). Moons are also typically much less massive than their planets, so a moon's orbit can be used to get the mass of its planet. Question 15 One of Jupiter's moons, Europa, orbits Jupiter with a period of 3.6 days (0.01 years) with a semi-major axis of 004 AU. Calculate the mass of Jupiter using the above equation (remember your units). e Question 16 Now, look at the mass (in solar masses) you calculated for Jupiter. Does this seem plausible? Why, or why not?

Answer 1

T: 3.6x24 3600 : 3.1x10 sec The Semr maon axis a-o-004Au )月u- l.yq620,n using keplen, S third law e-e 9-61x,。,。」专.SM2-rd', × 2.yrio26め3 on, 0几 37 461 x10

k. we found the sacon mos jupiten is. 00066 Small (o.oous) anQdhe Solon mass mo.a3 α- τ Yest can be Pbe on the iven value!