Question

1. the earth moves around the sun with a tangential velocity equal to roughly 30,000 m/s.

a. if we take the earths period to be exactly 365 days, what is its period in seconds?
b. calculate the distance between the earth and the sun.
c. earths mass is approximately 5.97 x 10^24 kg. use the centripetal force equation to calculate the gravitational force between earth and the sun.
d. use this gravitational force and distance to calculate the mass of the sun.
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Explanation:

Part a

Step1: Convert days to hours

Since \(1\) day \( = 24\) hours, for \(365\) days, the number of hours is \(365\times24\) hours.

$$365\times24=8760$$

Step2: Convert hours to minutes

Since \(1\) hour \( = 60\) minutes, for \(8760\) hours, the number of minutes is \(8760\times60\) minutes.

$$8760\times60 = 525600$$

Step3: Convert minutes to seconds

Since \(1\) minute \(=60\) seconds, for \(525600\) minutes, the number of seconds is \(525600\times60\) seconds.

$$525600\times60=31536000$$

Step1: Use the formula for circular motion

The formula for the circumference of a circle \(C = 2\pi r\), and also \(v=\frac{C}{T}\) (where \(v\) is tangential velocity, \(T\) is period). Rearranging for \(r\), we get \(r=\frac{vT}{2\pi}\).
We know \(v = 30000\space m/s\) and \(T=31536000\space s\).

$$r=\frac{30000\times31536000}{2\pi}$$

Step2: Calculate the value

$$r=\frac{9.4608\times 10^{11}}{2\pi}\approx1.5\times 10^{11}\space m$$

Step1: Use the centripetal force formula

The centripetal force \(F_c=\frac{mv^{2}}{r}\). Substitute \(m = 5.97\times 10^{24}\space kg\), \(v = 30000\space m/s\), and \(r=1.5\times 10^{11}\space m\)

$$F_c=\frac{5.97\times 10^{24}\times(30000)^{2}}{1.5\times 10^{11}}$$

Step2: Simplify the expression

$$F_c=\frac{5.97\times 10^{24}\times9\times 10^{8}}{1.5\times 10^{11}}=\frac{5.97\times9}{1.5}\times10^{24 + 8-11}$$

$$F_c = 3.582\times10^{22}\space N$$

Answer:

\(31536000\) seconds

Part b