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Question: The earth (mass = \({10^{24}}\)Kg) revolves around the sun with angular velocity \(2 \times {10^{ - ...

The earth (mass = 1024{10^{24}}Kg) revolves around the sun with angular velocity 2×10−7 rad s−12 \times {10^{ - 7}}{\text{ rad }}{{\text{s}}^{ - 1}} in a circular orbit of radius 1.5×108 Km1.5 \times {10^8}{\text{ Km}}. Find the force exerted by the sun on the earth (in × 1021{10^{21}}N).

Explanation

Solution

To find the force exerted by the sun on earth, we observe that it is nothing but the centripetal force required by the earth to revolve around the sun. We apply the formula of centripetal force. Centripetal force is defined as the force that aids a body move in a circular path. It always acts orthogonal to the direction of motion of the body.

Complete step by step answer:
Given Data,
Mass of earth = 1024{10^{24}}Kg
Angular velocity of earth ω = 2×10−7 rad s−12 \times {10^{ - 7}}{\text{ rad }}{{\text{s}}^{ - 1}}
Radius of the orbit of earth = 1.5×108 Km1.5 \times {10^8}{\text{ Km}}
We know 1 Km = 1000 m, therefore Radius of the orbit of earth = 1.5×1011 m1.5 \times {10^{11}}{\text{ m}}

Formula Used,
Centripetal force F = mrω2{\text{F = mr}}{\omega ^2}.
Centripetal force is defined as the force that is acted on the body moving in a circular path and it is directed towards the point with respect to which the body is rotated.
The centripetal force acting on a body of mass ‘m’ with angular velocity ‘ω’ moving in a circular path with radius ‘r’ is given by
F = mrω2{\text{F = mr}}{\omega ^2}
Therefore the centripetal force acting on earth is given by
⇒F = 1024×(1.5×108×103)×(2×10−7)2 ⇒F = 1024×(1.5×1011)×(4×10−14) ⇒F = 1024+11−14×1.5×4 ⇒F = 6×1021 N  \Rightarrow {\text{F = }}{10^{24}} \times \left( {1.5 \times {{10}^8} \times {{10}^3}} \right) \times {\left( {2 \times {{10}^{ - 7}}} \right)^2} \\\ \Rightarrow {\text{F = }}{10^{24}} \times \left( {1.5 \times {{10}^{11}}} \right) \times \left( {4 \times {{10}^{ - 14}}} \right) \\\ \Rightarrow {\text{F = }}{10^{24 + 11 - 14}} \times 1.5 \times 4 \\\ \Rightarrow {\text{F = 6}} \times {\text{1}}{{\text{0}}^{21}}{\text{ N}} \\\
Hence the force exerted by the sun on the earth is 6×1021 N{\text{6}} \times {\text{1}}{{\text{0}}^{21}}{\text{ N}}.

Note:
In order to answer this type of question the key is to know the phenomena of centripetal force and its formula.
A few examples of centripetal force phenomena in daily life are –
Spinning a rock tied to a string, here the string provides the centripetal force required to rotate the rock.
Turning a car on the road, here the friction between the road and the wheels provides the necessary centripetal force.
Centripetal force always acts towards the center of rotation, it pulls the body towards the center of its point of rotation.
It is important to convert all the given quantities into S.I units before substituting them in the formula. The S.I units of force is Newton ‘N’.