# Could someone show me how to do this formula?

Could someone use this formula to solve for Mars? This is the Period of a Planet Orbiting the Sun formula.

T = 2π√(r^3/Gm)

T = Period of Mars orbiting the Sun

r = radius

G = Universal Gravitational Constant

m = mass of the Sun

And also, what unit is T in? And could you show how the units cancel out to get T?

Thanks

I believe T has to be equal to around 686 days

r = 3.4 x 10^6

G = 6.67300 × 10^11 (N*m^2) / (kg^2)

m = 1.99 x 10^30

Thank you very much. It's a tough problem.

Oops, wrong number.

r = 2.28 x 10^11

Sorry... The other one was the radius of the plant mars. Not the orbital radius.

### 2 Answers

- JullyWumLv 79 years agoFavorite Answer
T = 2π √(r³ / Gm) ...... time always in seconds! ( as G is defined in terms of seconds)

doing the (r³/Gm) bit first ..

r³ = (2.28*10^11)³ = 1.185*10^34

Gm = 6.67*10^-11 * 1.99*10^30 = 1.327*10^20 .............(note G has 10^-11)

r³/Gm = 8.927*10^13

√(r³/Gm) = 9.45*10^6

► T = 2π x 9.45*10^6 = 5.94*10^7 seconds ( ÷ 3600 = hrs ... ÷ 24 = days → 687 days)

• To show the units cancelling down to T ..do you really want to do this !! ☺

Using notation M = mass (kg) .. L = length (m) .. T = time (s)

N (Newtons) = (m x a ) = ( M L T ̅ ² )

Gm =( Nm² / kg²) x m .. = ( M L T ̅ ² )( L²) (M ̅ ² ) x (M) = L³ T ̅ ²

r³/Gm = (L³) ÷ (L³ T ̅ ²) = (L³) x (L ̅ ³ T²) = T²

hence √(r³/Gm) = √ T² = T ☺

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- 9 years ago
I could solve this for you, but I would need the actual values for every magnitude. Also, the dimensional analysis (how the units cancel out) is easy to do but I need the units of G so I can work it out

Oh and by the way, T is measured in the regular units of time. You could use hours, minutes, days... The standard is seconds though.

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