Calculating the Period of Mars in Earth Years Using Gravitational Equations

[krimor09]

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

 

[just__curious]

any ideas?

 

[krimor09]

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

 

[just__curious]

Your approch is right but check your math. Note: you made 3 errors.

 

[jgens]

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.

 

[krimor09]

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

 

[jgens]

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.

 

[krimor09]

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

 

[jgens]

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

 

[krimor09]

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

 

[jgens]

Yes. Given your data, I would use 1.87 years (3 sig. figs.).

 

[krimor09]

thank you! i appreciate your patience

 

[jgens]

You're welcome! :)