Two bodies of equal masses revolve in circular orbits of radii R₁ and R₂ with the same period. Their centripet

Question

Two bodies of equal masses revolve in circular orbits of radii R₁ and R₂ with the same period. Their centripetal forces are in the ratio:

Accepted Answer

1. **Identify the Given Conditions:**
   - Masses are equal ($m_1 = m_2 = m$).
   - Time periods are equal ($T_1 = T_2 = T$).
   - Radii are $R_1$ and $R_2$.

2. **Select the Relevant Formula:**
   The centripetal force ($F_c$) required for a body of mass $m$ moving in a circle of radius $R$ is given by $F_c = m a_c$. Since the time period ($T$) is given, it is convenient to use the angular velocity form of centripetal acceleration, $a_c = \omega^2 R$, where $\omega = \frac{2\pi}{T}$.
   $$ F_c = m \omega^2 R = m \left(\frac{2\pi}{T}\right)^2 R $$
   [Source 47]

3. **Analyze the Proportionality:**
   Since $m$ and $T$ (and thus $\omega$) are the same for both bodies, the term $m(2\pi/T)^2$ is a constant constant ($k$).
   $$ F_c = k R $$
   Therefore, the centripetal force is directly proportional to the radius ($F_c \propto R$).

4. **Calculate the Ratio:**
   $$ \frac{F_1}{F_2} = \frac{R_1}{R_2} $$

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