3.1 - Kepler & Newton

[@lunattic22]

At min 14:30.. if 'n' is the satellite, and 1 is the Earth... what about 2? what is 'm2' ?

[@jadegecko]

m1 + m2 describes the masses of the two objects - in this case m2 would be the mass of the satellite. However, in the case of something as big as the earth vs something as small as an artificial satellite, m2 is negligible.

[@volleysmackz5960]

@@jadegecko But the point is satellite under test is referenced as nth object, why are we considering the mass of subject under test while calculating it's acceleration?

[@JDechnics]

@@volleysmackz5960 I think this is a mistake again.

[@李炆芮]

it must be satellite, the first half shows the earth's effect on the satellite

[@coreC..]

It is rather confusing. First you see that equation (calculating acceleration of nth body) where the nth body is not used in the sum (j=1 to J), and next is that equation (of the same acceleration of nth body) with a split-up version on the right. The gentleman clearly says: "think the body under test, body n, is the satellite". "Think of body 1 as being the Earth". The part on the very right side of the equation (only that sum j=3 to J) "are due to the moon, and the planets and the sun". => m2 must be the mass of the satellite. Which makes that next equation differ (including body n) from that first equation (excluding body n). However, i must agree with "jadegecko" in his statement, because: The gravitational force due to the Earth on the satellite, is equal to the force due to the satellite on the Earth. However, that mass of the satellite is so small, it can practically be ignored. If you ignore that m2 mass, you make the lower equation the same as the upper one, by excluding that mass of the nth body. But still, the equations are not the same, but the calculations will yield the same answer.

[@Jordatech]

Excellent ! One minor units correction, G = 6.6*10^-11 m^3 kg^-1 s^-2.

[@coreC..]

Good eye.