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?
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.
- 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 ☺
- 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.Source(s): Engineering student