Mass of Planet X in Solar Masses

AI Thread Summary
Planet X has a small moon that orbits it monthly at a distance of 15 times its diameter, which is 0.012 AU. Using Newton's form of Kepler's Third Law, the orbital period (P) is 30 days. The mass of Planet X can be calculated with the formula M = a^3/P^2, where 'a' is the semi-major axis in astronomical units. The calculations initially led to confusion, but the correct approach was confirmed. Ultimately, the discussion highlights the application of Kepler's laws to determine the mass of celestial bodies.
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Planet X is observed to have a small moon. This moon is observed to orbit the planet once per month at a distance of 15 Planet X diameters. What is the mass of Planet X in solar masses? [Hint: Use Newton's form of Kepler's Third Law: P2 is proportional to a3/M, where M is the mass of Planet X (assuming that the mass of the moon is negligible in comparison).]

the diameter of planet X is .0008 au as derived from previous problems. so 15 planet x diameters would be .012

So this would be a correct?
And p would be 30 days?

so the mass would be M = .012^3/30^2?

If this is correct am i missing something because i cannot come up with the correct anwer.
 
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figured it out... thanks anyway
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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