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## GR0877 Problem 53

PROBLEM STATEMENT: This problem is still being typed.

SOLUTION: (D) According to the problem statement, the angular momenta are the same: $m_1 v_1 R = m_2 v_2 R$, therefore $m_1/m_2 = v_2/v_1$. For the orbital periods one has $T_1 = 2\pi R/v_1$ and $T_2 = 2 \pi R/v_2$ (orbits are circles!). Thus, $m_1/m_2 = v_2/v_1 = T_1/T_2 = 3$. Think about why cannot we apply Kepler’s third law in its usual form.

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1. 05.03.2014 at 06:40

The main reason we can’t apply Kepler’s third law is that the corresponding masses of their suns are different.

2. 01.04.2014 at 05:03

Why isn’t it stated in the question that two solar system has different solar mass? At first I thought they meant they observed two suns, like two stars with identical solar mass of 1.

3. 24.10.2014 at 06:43

It’s stupidly worded. Thanks again, ETS. You’re always so clear and concise.

4. 29.08.2016 at 09:20

Alternatively, remember that L=m*(dA/dt), where dA/dt is the rate at which area swept out by the planet. Integrate both sides from t=0 to t=T to get that LT=mA, or rather, m=LT/A. This says that the ratio (m1/m2) is equal to (L1*T1/A1)/(L2*T2/A2). But L1=L2 and A1=A2, so (m1/m2)=(T1/T2).