The earth mass-6- 1024kg revolves round the sun with angular velocity 2- 10-7rad-s in a circular orbit of radius 1-5- 108 km- The force exerted by the sun on the earth in newtons- is

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Standard XIII

Physics

Newton's Law of Gravitation

The earth mas...

Question

The earth $(m a s s = 6 \times 10^{24} k g)$ revolves round the sun with angular velocity $2 \times 10^{- 7} r a d / s$ in a circular orbit of radius $1.5 \times 10^{8} k m$ . The force exerted by the sun on the earth in newtons, is

$18 \times 10^{25}$

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Zero

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$27 \times 10^{39}$

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$36 \times 10^{21}$

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Solution

The correct option is D
$36 \times 10^{21}$

$m = 6 \times 10^{24} k g, ω = 2 \times 10^{- 7} r a d s, R = ¯ ¯¯¯¯¯ ¯ 1.5 \times 10^{11} m$
The force exerted by the sun on the earth $F = m ω^{2} R$
By substituting the value we can get, $F = 36 \times 10^{21} N$

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Q. The earth

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revolves around the sun with angular velocity

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km. The force exerted by the sun on the earth in newtons is:

The earth $(m a s s = 6 \times 10^{24} k g)$ revolves around the sun with angular velocity $2 \times 10^{- 7} r a d / s$ in a circular orbit of radius $1.5 \times 10^{8} k m$ . The force exerted by the sun on the earth in newtons is

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The earth (mass=6×1024kg) revolves round the sun with angular velocity 2×10−7rad/s in a circular orbit of radius 1.5×108 km. The force exerted by the sun on the earth in newtons, is

The correct option is D 36×1021 m=6×1024kg,ω=2×10−7rads,R=¯¯¯¯¯¯¯1.5×1011m The force exerted by the sun on the earth F=mω2R By substituting the value we can get, F=36×1021N

The earth $(m a s s = 6 \times 10^{24} k g)$ revolves round the sun with angular velocity $2 \times 10^{- 7} r a d / s$ in a circular orbit of radius $1.5 \times 10^{8} k m$ . The force exerted by the sun on the earth in newtons, is

The correct option is D
$36 \times 10^{21}$

$m = 6 \times 10^{24} k g, ω = 2 \times 10^{- 7} r a d s, R = ¯ ¯¯¯¯¯ ¯ 1.5 \times 10^{11} m$
The force exerted by the sun on the earth $F = m ω^{2} R$
By substituting the value we can get, $F = 36 \times 10^{21} N$