Calculating the Period of Mars in Earth Years Using Gravitational Equations

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Homework Help Overview

The discussion revolves around calculating the orbital period of Mars in Earth years using gravitational equations, specifically Kepler's third law. Participants are analyzing the relationship between the distances of Earth and Mars from the Sun and their respective periods.

Discussion Character

  • Mathematical reasoning, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the application of Kepler's third law, attempting to set up equations relating the periods and distances of Earth and Mars. There are multiple attempts to manipulate the equations and check calculations, with some participants expressing uncertainty about their errors.

Discussion Status

The discussion is ongoing, with participants providing feedback on each other's calculations and approaches. Some guidance has been offered regarding the need to convert time into years and to double-check mathematical operations. There is no explicit consensus on the correct period yet, as various interpretations and calculations are being explored.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the amount of assistance they can receive. There is an emphasis on checking calculations and ensuring proper unit conversions, particularly from days to years.

krimor09
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Homework Statement


Distance of Earth from Sun = 1.50 x 10^11m
Period = 365.2 days
Distance of Mars from Sun = 2.28 x 10^11km
Period of Mars in Earth Years?


Homework Equations


(Ta/Tb)^2 = (Ra/Rb)^3


The Attempt at a Solution



(365.2 days/Tm)^2 =(1.50 x 10^11m/2.28x10^14m)^3
=(365.2/Tm)^2 = 2.85 x 10^74

Tm^2= (365.2)^2 (2.85 x 10^74)
= square root of 3.04 x 10^79
Tm = 5.5 x 10^39
 
Last edited:
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any ideas?
 
i just put up my work, i know that its wrong because my physics teacher said so, but I'm not sure what my major error is
 
Your approch is right but check your math. Note: you made 3 errors.
 
Several errors. First you may want to rewrite your equations (and double check your calculations). Second, you need to convert your time into years.

Start here (t/1 yr)^2 = (2.28 x 10^11/1.50 x 10^11)^3.
 
okay, i tried this way out

(t/365.2days)^2 = (2.28x10^11/1.50x10^11)^3
(T/365.2) = square root of 3.5
T^2/133371 = 1.87
multiply 133371 to both sides
t^2 = the square root of 1.87 X 133371
T = 499
 
Again, you need to check your calculations (and convert your time into years!). When you take the sqrt of 3.5 you have (t/365.2) = 1.87, not (t/365.2)^2 = 1.87. Be careful with the algebra.
 
Dm= 2.28 x 10^8m
De= 1.50 x 10^11m
Te= 365.2 days or 1 year
Tm= ?

(T/1 yr)^2 = (2.28 x 10^8m/1.50 x 10^11m)^3
(T^2/1yr) = 3.5 x 10^57)
T^2 = sqrt of 3.5 x 10^57
= 5.9 x 10^28
 
Again, incorrect. Be careful with calculations and what numbers you substitute into the equation. Your first equation should read:

t^2 = (2.28 x 10^11/1.50 x 10^11)^3
 
  • #10
t^2 = (2.28 x 10^11/1.50 x 10^11)^3
t^2 = sqrt of 3.5
t = 1.87 years or about 2 years
 
  • #11
Yes. Given your data, I would use 1.87 years (3 sig. figs.).
 
  • #12
thank you! i appreciate your patience
 
  • #13
You're welcome! :)
 

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