The Earth takes one year to orbit the Sun at a distance of 1.5 x 10^11 m. Calculate i

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Swoosh
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The Earth takes one year to orbit the Sun at a distance of 1.5 x 10^11 m. Calculate its speed.


Speed = distance/time

(1.5 x 10^8)/ (365*24*60*60) = 4.75646... kms-1




Why is the answer in the book 29.9 kms-1?
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DumpOrStay?
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Because your title has to the power 11 and your working has to the power 8, maybe?

Either work in metres or km.
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didgeridoo12uk
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distance is wrong. you've used the radius not circumference.

basically you've calculated what the average speed of the earth would be if it took a year to crash directley into the sun
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tieyourmotherdown
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Like, really really fast.
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Swoosh
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(Original post by DumpOrStay?)
Because your title has to the power 11 and your working has to the power 8, maybe?

Either work in metres or km.
I changed the distance to km because the answer in the book is given in km. If I do the same sum with x10^11 it's just 1000 x bigger :confused:
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Dynasty
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i = \sqrt -1
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Swoosh
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(Original post by didgeridoo12uk)
distance is wrong. you've used the radius not circumference.

basically you've calculated what the average speed of the earth would be if it took a year to crash directley into the sun
Ah ok.

So the distance given is the radius? So the correct sum would be


(2 x pi x 1.5x10^8) / (365*24*60*60)

= 29.8857...


Yay

Thanks
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Steven
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The motion is circular, so you must find the circumference of the orbit first.

Pi x diamter = 3.14 x (2x1.5x10^8)=9.24x10^8.

Then use speed = distance/time

Speed = 9.24x10^8) / (365x24x60x60)=29.885 kms^-1
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eba sherin
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becos its the radius u need to find the circumference of the earth 2*pi*r
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