Problem Statement
Solve the Newton’s Laws / mechanics problem: Solve the Newton’s Laws / mechanics problem: Earth orbits the Sun at radius $r = 1.5\times10^{11}$ m with period $T = 3.15\times10^7$ s. Find the centripetal acceleration. ($M_{\odot} = 2\times10^{30}$ kg, $G = 6.67\times10^{-11}$) $a_c = v^2/r = (2\pi r/T)^2/r = 4\pi^2 r/T^2$ Step 1: $a_c = 4\pi^2
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
Circular motion requires a centripetal force directed toward the centre, providing the centripetal acceleration $a_c = v^2/r = \omega^2 r$. This force is not a new type of force — it is always the resultant of real forces (tension, normal force, friction, gravity) directed inward. At the minimum speed for maintaining contact, the normal force drops to zero.
- $a_c = v^2/R = \omega^2 R$ — centripetal acceleration
- $F_c = mv^2/R$ — net centripetal force needed
- Banked curve: $\tan\theta = v^2/(Rg)$ — ideal banking angle
- Loop minimum speed: $v_{\min} = \sqrt{gR}$ at top (N=0)
Step-by-Step Solution
Step 1 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 2 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 3 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Worked Calculation
Full substitution shown in the steps above.
Answer
$$\boxed{a = \dfrac{(m_2-m_1)g}{m_1+m_2}}$$
Physical Interpretation
The centripetal force is not a ‘new’ force but the net inward resultant of real forces. If that resultant falls below $mv^2/r$, the object cannot maintain circular motion and will fly outward — this is the critical condition for minimum speed problems.
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