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nrlwinner

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Hi. If anyone has the answers to "28. Kepler's Law of Motion" in the Surfing Physics Space book, could they please post it up here so I can check my answers.

Thanks in advance.
 

Aerath

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If you type out the question, I can tell you if I get the same answer. [I don't have the book]
 

nrlwinner

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Nah there's 10 questions in the set.

But if you still want me to type it up, I'll do that.
 

nrlwinner

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I'll type up the questions I'm pretty sure is wrong.

3) Two minor planets of the Solar System have periods of 3.0 years and 4.0 years. What is the ratio of the average radii of their orbits.

4) On a distant solar system, two planets were observed to have radii of 4AU and 6AU. What would be the ratio of the periods of revolution of the two planets? (Note: 1AU is the radius of the orbit of the Earth around our Sun)

9) Using the data of Q8, calculate the height of a space station above the Earth's surface which would have an orbital period of 90 days. The radius of the Earth is 6.38 x 10^6 km.

(Question 8 is: Considering the radius of the Moon's orbit around the Earth is 384405km and its period is 27.322 days, what would be the orbital radius of a satellite have to be in order for it to remaine above the same geographic point on the Earth's equator given that the Earth rotates once in 23h 56 min
I got the answer to be: 42296.2 km)

10) When it is time for a space shuttle from the space station to return to Earth, the shuttle slows down until its orbit takes it it the upper reaches of the atmosphere, approxiamately 100km abover the surface of the Earth. Calculate the orbital period of the shuttle at this stage and its linear velocity.



I've done every question, but for these ones I'm pretty sure my answer is wrong.
 

Aerath

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Physics was my worst subject at school, so these are probably wrong, but here we go. :p

 

adomad

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9) Using the data of Q8, calculate the height of a space station above the Earth's surface which would have an orbital period of 90 days. The radius of the Earth is 6.38 x 10^6 km.

(Question 8 is: Considering the radius of the Moon's orbit around the Earth is 384405km and its period is 27.322 days, what would be the orbital radius of a satellite have to be in order for it to remaine above the same geographic point on the Earth's equator given that the Earth rotates once in 23h 56 min
I got the answer to be: 42296.2 km)



I've done every question, but for these ones I'm pretty sure my answer is wrong.
the answer for 8 is 376684.7874km (minus the radius of the Earth). just a tip you should be remembering that a geostationary satellite has a altitude of 385 000 km so 42 296.2 km would be too small...

going on to Q9 ( i don't think u need any of the data from 8)

using Kepler's laws.... T= 90*24*60*60 R= ? G= 6.67x10^-11 M = 5.97x10^24

.'. R^3 ={(90*24*60*60)^2 * 6.67x10^-11 * 5.97x10^24) / (4pi^2)
.'. R=841 669.5516 km ( i took the Earths radius to be 6373x10^3 (force of habit)

for Q10.

use Kepler's laws and to find the orbital velocity using the formula v= *square root* [(Gm)/r]

i get the period to be 1.44 hours and the orbital velocity to be 7843.24ms-1 ( i don't think that the orbital speed looks right lol just taking a punt)
 
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