Question

when and why should i convert T into seconds?

$352833c3fe6d3ac93-46b2263058d8b53157platforms_id=mastering&b-HOIM2W/BBP&idpName=SMS&.contextid=UIC PHYSICS141FALL20198 smsUse 13.5: Kepler's Laws And The Motion of Planets Example 13.8 Kepler's third law WITH VARIATION PROBLEMS The asteroid Pallas has an orbital period of 4.62 years and an orbital eccentricity of 0.233. Find the semi-major axis of its orbit. IDENTIFY and SET UP We need Kepler's third law, which relates the period T and the semi-major axis a for an orbiting object (such as an asteroid). We use Eq. (13.17) to determine a; from Appendix Fo we have ms -1.99 x 10 kg, and a conversion factor from Appendix ED gives T = (4.62 y)(3.156 x 10's/y) - 1.46 x 10Ⓡs. Note that we don't need the value of the eccentricity. EXECUTE From Eq. (13.17), 23/2 = (Gms.)/2T]/2*. To solve for a, we raise both sides of this expression to the power and then substitute the values of G, ms, and T: a= (mr) – 4.15 x 10 m Gma 721/3 4,2 (Plug in the numbers yourself to check.) EVALUATE Our result is intermediate between the semi-major axes of Mars and Jupiter (see Appendix FO). Most known asteroids orbit in an "asteroid belt between the orbits of these two planets. KEY CONCEPT Kepler's third law allows you to relate the period of an orbit, the semi-major axis of that orbit, and the mass of the object being orbited. Video Tutor Solution: Example 13.8

Answer 1

The SI unit of universal gravitational constant is m^3/(kg.s^2).

So we have to use other values in same units to calculate our result in SI unit.

So we put value of Time = T in seconds only which is given T=1.46*10^8 sec.

So as to get our answer in meter.