More practice questions

4. Q4. The Moon goes round the Earth once in 27.3 days. Treating its orbit as a circle, find its radius. Take g = 9.8 m s^⁻² and R_E = 6.4 × 10^⁶ m.

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   r^³ = gR_E²T^²/(4π^²) with T = 27.3 × 86 400 s, about 2.36 × 10^⁶ s. This gives r ≈ 3.84 × 10^⁸ m, about 60 Earth radii.

5. Q5. Show that a satellite skimming the surface of any uniform planet of density ρ has a period T₀ = √(3π/(Gρ)), whatever the planet’s size. Use T₀ = 84.6 minutes to find the Earth’s mean density.

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   T₀² = 4π^²R^³/(GM) and M = (4/3)πR^³ρ, so T₀² = 3π/(Gρ); R cancels. With T₀ = 5076 s, ρ = 3π/(GT₀²), about 5.48 × 10^³ kg m^⁻³.

6. Q6. NCERT Example 7.7 (page 138): express k = 10^⁻¹³ s^² m^⁻³ in days and kilometres, and use it to find the Moon’s period, its distance being 3.84 × 10^⁵ km.

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   Since 1 s = 1/86 400 d and 1 m = 10^⁻³ km, 1 s^² m^⁻³ = (1000)^³/(86 400)^² d^² km^⁻³, so k = 10^⁻¹³ × (1000)^³/(86 400)^², about 1.34 × 10^⁻¹⁴ d^² km^⁻³ (NCERT prints 1.33 × 10^⁻¹⁴). Then T^² = 1.34 × 10^⁻¹⁴ × (3.84 × 10^⁵)^³, about 759, so T ≈ 27.5 days. NCERT prints 27.3 days, the Moon’s measured period, but its numbers give about 27.4 to 27.5 days; k = 10^⁻¹³ s^² m^⁻³ is itself rounded from about 9.9 × 10^⁻¹⁴ s^² m^⁻³.

7. Q7. If a satellite’s orbit radius is increased by 1%, by roughly what percentage does its period change?

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   T ∝ r^3/2, so a small fractional change in r gives 3/2 times that change in T: the period grows by about 1.5%.