Earth's Average Speed: Calculating its Orbit around the Sun

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The Earth orbits the sun in a nearly circular path with a distance of approximately 1.50 x 10^11 meters. Over the summer months, which span about 7.89 x 10^6 seconds, the Earth covers a quarter of its orbital distance. To calculate the average speed, one can use the formula average speed = distance/time taken, incorporating the necessary mathematical constants. The discussion humorously notes the need for "a slice of pi," indicating the relevance of pi in circular motion calculations. Overall, the conversation emphasizes the straightforward approach to determining Earth's average orbital speed.
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The Earth moves around the sun in a nearly circular orbit of 1.50 x 10m ^ 11

During the three summer months (an elasped time of 7.89 x 10s^6) the Earth moves 1/4 the distance around the sun.

What is the average speed of the Earth?
 
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If this is a homework question it should be posted in the homework forum, i think someone will come and move it.

Well if you use average speed = distance/time taken you will come up with the answer.
 
You will also need a slice of pi.
 
Chi Meson said:
You will also need a slice of pi.
mmmm! pi!
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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