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Chapter 6: Problem 675

The earth (mass \(=6 \times 10^{24} \mathrm{~kg}\) ) revolves around the sun with angular velocity \(2 \times 10^{-7} \mathrm{rad} / \mathrm{sec}\) in a circular orbit of radius \(1.5 \times 10^{8} \mathrm{~km} .\) The force exerted by the sun on the earth is \(=\ldots \ldots \ldots \ldots . \mathrm{N}\) (A) \(18 \times 10^{25}\) (b) zero (C) \(27 \times 10^{39}\) (D) \(36 \times 10^{21}\)

Short Answer

Expert verified
The force exerted by the sun on the earth is (D) \(36 \times 10^{21} N\).

Step by step solution

01

Convert the radius to meters

Given the radius in kilometers, we need to convert it to meters first. To convert the radius from kilometers to meters, we can multiply by \(1,000\). \[r = 1.5 \times 10^8 km \times 1,000 = 1.5 \times 10^{11} m\]
02

Calculate linear velocity (v)

Next, we need to calculate the linear velocity using angular velocity and radius: \[v = r\omega\] \[v = (1.5 \times 10^{11} m) (2 \times 10^{-7} \frac{rad}{s})\] \[v = 3 \times 10^4 \frac{m}{s}\]
03

Calculate centripetal force (F)

Now, using the centripetal force formula, we can find the force exerted by the sun on the earth: \[F = m \frac{v^2}{r}\] \[F = (6 \times 10^{24} kg) \frac{(3 \times 10^4 \frac{m}{s})^2}{1.5 \times 10^{11} m}\]
04

Simplify the expression

Let's simplify the equation to find the force: \[F = (6 \times 10^{24} kg) \frac{(9 \times 10^8 \frac{m^2}{s^2})}{1.5 \times 10^{11} m}\] \[F = (6 \times 10^{24} kg) (6 \times 10^{-3} \frac{m}{s^2})\]
05

Calculate the force

Finally, let's calculate the force exerted by the sun on the earth: \[F = 36 \times 10^{21} N\] From the given options, the correct answer is (D) \(36 \times 10^{21} N\) .

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