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Earth’s orbits once around the sun every 365.25 days at an average radius of 1.5x10^11m. Earth’s mass is 6x10^24kg. a) what distance does earth travel in 1 year?b) what is earths average centripetal acceleration?c) what is the average force that the sun exerts on earth?

Answer :

Answer:

distance = 9.42 x 10^11 m

Acceleration = 5.94 x 10^-3 m/s2

Force = 3.56 x 10^22 N

Explanation:

Part a)

The distance travel by the earth in 1 year can be calculated as the circumference of a circle of radio r, so:

[tex]\begin{gathered} d=2\pi r \\ d=2(3.14)(1.5\times10^{11}m) \\ d=9.42\times10^{11}m \end{gathered}[/tex]

So, the distance is 9.42 x 10^11 m

Part b)

The centripetal acceleration can be calculated as:

[tex]a_c=\frac{v^2}{r}[/tex]

Where v is the velocity and r is the radius of the earth. So, the velocity is equal to:

[tex]v_{}=\frac{\text{distance}}{\text{time}}=\frac{9.42\times10^{11}m}{365.25\text{days x 24 hr x 60 min x 60 s}}=29850.18\text{ m/s}[/tex]

Therefore, the centripetal acceleration is:

[tex]a_c=\frac{(29850.18)^2}{1.5\times10^{11}}=5.94\times10^{-3}m/s^2[/tex]

Part c)

Finally, the average force that the sun exerts on the earth can be calculated by the second law of Newton as:

[tex]\begin{gathered} F=m\cdot a_c \\ F=(6\times10^{24})\cdot(5.94\times10^{-3}) \\ F=3.56\times10^{22}\text{ N} \end{gathered}[/tex]

Therefore, the answers are:

distance = 9.42 x 10^11 m

Acceleration = 5.94 x 10^-3 m/s2

Force = 3.56 x 10^22 N