HC Verma Chapter 7 Problem 29 — Centripetal force in planetary orbit

Problem Statement

Solve the Newton’s Laws / mechanics problem: Solve the Newton’s Laws / mechanics problem: Earth (mass $6\times10^{24}$ kg) orbits the Sun at $r = 1.5\times10^{11}$ m with $T = 1$ year $= 3.15\times10^7$ s. Find the centripetal force on Earth. $F_c = m\omega^2 r = m(2\pi/T)^2 r$ Step 1: $\omega = 2\pi/(3.15\times10^7) = 1.995\times10^{-7}$ rad/

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

$$\sum F_x = ma_x\quad,\quad \sum F_y = ma_y = 0\quad\text{(if no vertical acceleration)}$$

$$a = \frac{(m_2-m_1)g}{m_1+m_2} = \frac{(5-3)\times9.8}{8} = \frac{19.6}{8} = 2.45\,\text{m/s}^2$$

$$T = \frac{2m_1 m_2 g}{m_1+m_2} = \frac{2\times3\times5\times9.8}{8} = \frac{294}{8} = 36.75\,\text{N}$$

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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