Question

1. neptune has a mass of 1.02 x 10²⁶ kg, and a radius of 2.48 x 10⁷ m. assuming you could stand on the surface of neptune, what gravitational field strength would you experience?
2. 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 10²⁴ 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.

Explanation:

8a.

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: Recall the formula for circumference

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

Step2: Substitute values into the formula

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

$$r=\frac{9.4608\times 10^{11}}{2\pi}$$

$$r\approx1.506\times 10^{11}\text{ m}$$

Step1: Recall the centripetal force formula

The centripetal force formula is \(F = m\frac{v^{2}}{r}\).
Given \(m = 5.97\times 10^{24}\) kg, \(v = 30000\) m/s, and \(r\approx1.506\times 10^{11}\) m.

Step2: Substitute values into the formula

$$F=(5.97\times 10^{24})\frac{(30000)^{2}}{1.506\times 10^{11}}$$

$$F=(5.97\times 10^{24})\frac{9\times 10^{8}}{1.506\times 10^{11}}$$

$$F=\frac{5.97\times9\times 10^{24 + 8}}{1.506\times 10^{11}}$$

$$F=\frac{53.73\times 10^{32}}{1.506\times 10^{11}}$$

$$F\approx3.57\times 10^{22}\text{ N}$$

Answer:

\(31536000\) seconds

8b.