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Q7.3: NCERT Physics - Part 1 GRAVITATION 7.3 Solution: Suppose there existed a planet that went around the Sun twice…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently confuse the exponents in Kepler's third law, mistakenly equating the cube of the radius ratio to the time period ratio instead of its square. Always verify that the relationship is $(R_p/R_e)^3 = (T_p/T_e)^2$ to correctly handle the 'twice as fast' condition.

Suppose there existed a planet that went around the Sun twice as fast as the Earth. What would its orbital size be compared to that of the Earth?

Step-by-Step Verified Solution

**Answer:** The orbital radius of the planet would be (1/4)1/3≈0.63(1/4)^{1/3} \approx 0.63 times that of the Earth. **Step-by-Step Explanation:** Time period of Earth, Te=1T_e = 1 year. Time period of planet, Tp=0.5T_p = 0.5 year. Orbital radius of Earth, Re=1R_e = 1 AU. Orbital radius of planet =Rp= R_p. Using Kepler's third law:
(RpRe)3=(TpTe)2\left(\frac{R_p}{R_e}\right)^3 = \left(\frac{T_p}{T_e}\right)^2
(RpRe)3=(0.51)2=0.25\left(\frac{R_p}{R_e}\right)^3 = \left(\frac{0.5}{1}\right)^2 = 0.25
RpRe=(0.25)1/3≈0.63\frac{R_p}{R_e} = (0.25)^{1/3} \approx 0.63
The planet's orbital radius is approximately 0.63 times the Earth's orbital radius [citation:1].

All Questions in this Chapter

Q7.1 Q7.2 Q7.3 Q7.4 Q7.5 Q7.6 Q7.7 Q7.8 Q7.9 Q7.10 Q7.11 Q7.12 Q7.13 Q7.14 Q7.15 Q7.16 Q7.17 Q7.18 Q7.19 Q7.20 Q7.21

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