The earth (mass = 6 × 10²⁴ kg) revolves around the sun with an angular velocity of 2 × 10⁻⁷ rad/s in a circular orbit of radius 1.5 × 10⁸ km. The force exerted by the sun on the earth, in newton, is (a) 36 × 10²¹ (b) 27 × 10³⁹ (c) zero (d) 18 × 10²⁵

Answer: (a). Sun's pull supplies the centripetal force: F = mω² r. F = 6×10²⁴ × (2×10⁻⁷)² × 1.5×10¹¹ F = 6 × 4 × 1.5 × 10²⁴⁻¹⁴⁺¹¹ = 36 × 10²¹ N So the answer is (a) 36×10²¹.

The earth (mass = 6 × 10²⁴ kg) revolves around the sun with an… | NEET 1995 (AIPMT)

Q30NEET 1995 · AIPMTEasy

Kepler's Laws & Planetary Motion

The earth (mass $= 6 \times 10^{24}\text{ kg}$ ) revolves around the sun with an angular velocity of $2 \times 10^{-7}\text{ rad/s}$ in a circular orbit of radius $1.5 \times 10^{8}\text{ km}$ . The force exerted by the sun on the earth, in newton, is

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Answer: (a)

Sun's pull supplies the centripetal force: $F = m\omega^2 r$ .
$F = 6\times10^{24} \times (2\times10^{-7})^2 \times 1.5\times10^{11}$
$F = 6 \times 4 \times 1.5 \times 10^{24-14+11} = 36 \times 10^{21}\text{ N}$
So the answer is (a) $36\times10^{21}$ .

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