anonymous
  • anonymous
Two planets - planet A and planet B - are orbiting a star.If planet A has an orbital radius which is four times as large as planet B then the period of planet A orbit is _ times larger than the period of Planet B orbit
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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jchick
  • jchick
Applying Kepler's law of harmonies to this situation would result in: TA2 / RA3 = TB2 / RB3 This equation can be algebraically rearranged to TA2 / TB2 = RA3 / RB3 The ratio of the period squared of planet A to planet B will be equal to the ratio of the radius cubed of planet A to planet B. The ratio of the radii of the two planets is given - planet A's radius is two (or three) times larger than planet B's radius. The cube of this ratio is equal to the square of the ratio of the period. Taking the square root of the period squared ratio will yield the ratio of the periods of the planets. Mathematically, this could be written as TA / TB = SQRT(TA2 / TB2) = SQRT(RA3 / RB3)
anonymous
  • anonymous
answer 8
anonymous
  • anonymous
can u plug in the number

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anonymous
  • anonymous
can u help me plug in the numbers
jchick
  • jchick
@pooja195 I am blanking out on this can you help?
jchick
  • jchick
@Loser66 can you help I am blanking out.
pooja195
  • pooja195
@zepdrix
jchick
  • jchick
Ta2 / Tb2 = Ra3 / Rb3
jchick
  • jchick
Sorry I am trying to remember.
anonymous
  • anonymous
i know the answer is 8 but i cant put it in the formula
jchick
  • jchick
Wait how do you know the answer is 8 if you don't know how to input it?
anonymous
  • anonymous
i didnt use the formula thats why i tried doing 4^3=sqrt64 which equals 8
jchick
  • jchick
I can't remember sorry I wouldn't want to give you the wrong information.
anonymous
  • anonymous
can u help me understand the formula
jchick
  • jchick
T2 = kr3 . In every day terms, T is the time of the orbital period of the planet, K is a constant, and r is the ratio of the Sun to the planet. If more than on planet is involved, the ratio becomes (Ta2 / Tb2 ) = (Ra3 / Rb3 ). This is still used today to find distances of objects found inside the Milky Way.
anonymous
  • anonymous
can u plug it in
jchick
  • jchick
Can you tell me what eight is?
anonymous
  • anonymous
eight is the answer
jchick
  • jchick
Ok answer to what. The whole problem? Because I cannot give you straight answers I want you to help come to the answer so that you understand it.
jchick
  • jchick
Where did you get the answer?
anonymous
  • anonymous
i did this √4³ = 8
anonymous
  • anonymous
this is the answer to the whole problem but i dont know how to put it into your formula
jchick
  • jchick
Are you working with Kepler?
anonymous
  • anonymous
yes
anonymous
  • anonymous
TA / TB = SQRT(TA2 / TB2) = SQRT(RA3 / RB3) is this the formula
jchick
  • jchick
Yes now do you know what the satellite mean orbital radius is?
anonymous
  • anonymous
planet a orbital radius is 4 times as larger as planet B
jchick
  • jchick
Ok so what is planet B
jchick
  • jchick
What is the planetary mass?
anonymous
  • anonymous
2 pie idk
jchick
  • jchick
Here is a file that will help
1 Attachment
anonymous
  • anonymous
By performing simple mathematics for planet A and B T(A)^2 / T(B)^2 = R(A)^3 / R(B)^3 Given R(A) = 4X R(B) T(A)^2 / T(B)^2 = 64 (As 4X4X4=64) T(A) / T(B) = 8 Thus Planet A takes 64 takes 8 times of Time Period over Planet B so answer is 8 times
jchick
  • jchick
Correct how did you find it?
anonymous
  • anonymous
looked at the previous formula and assume the radius
jchick
  • jchick
Nice!
jchick
  • jchick
Did you see my file though?
anonymous
  • anonymous
yes thanks for the help
jchick
  • jchick
No problem sorry I couldn't help more.

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