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#centripetal force

11 public questions tagged with this topic.

In Rutherford’s model, what provides the centripetal force for an electron orbiting the nucleus in a hydrogen atom?

**Thomson's plum pudding model** positive charge uniformly distributed in sphere with electrons embedded, positive charge spread, fails to explain large angle scattering observed. Bohr's model introduces stationary orbits with quantized angular momentum L = n h/2π, physical basis de Broglie standing wave condition 2πr = n λ, circumference fits n wavelengths, explains line spectrum. The electrostatic force between the electron and nucleus provides the centripetal force: (e²/4πepsilon₀ r²) = (m v²/r) . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1 u = 931.5 MeV

Ref: NCERT > Physics Book > Atoms and Nuclei > Atomic Models - Rutherford, Thomson and Bohr

What is the primary source of centripetal force for an Earth satellite?

For a satellite in circular orbit, the centripetal force (F = mv2r) is provided by Earth’s gravitational force (F = GMEmr2), which keeps it in orbit. As per NCERT, applying relevant law/formula with correct units and sign convention leads to Gravitational force. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: Halliday & Resnick, Fundamentals of Physics, 11th ed., Chapter 13: Gravitation.

Why does the gravitational force provide the necessary centripetal force for a circular orbit?

In a circular orbit, the gravitational force (GMEmr2) equals the centripetal force (mv2r), providing the inward force needed to maintain circular motion. As per NCERT, applying relevant law/formula with correct units and sign convention leads to It balances the required inward force. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: Halliday & Resnick, Fundamentals of Physics, 11th ed., Chapter 13: Gravitation.

A 1000kg car on a banked road (θ\=15∘, μs\=0.3) turns at radius 40m. What is the maximum speed without slipping? (Take g

Max speed vmax=rgμs+tan⁡θ1−μstan⁡θ. As per NCERT Chapter 3, equations v=u+at, s=ut+½at², v²=u²+2as describe uniformly accelerated motion. Applying correct signs and units gives 14 m/s as the result, so option A is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Straight Line Motion - Practice Set0