Talk:Free-fall time

I suspect that in the following equation[1] someone messed up with the relation between orbital period, and the distance between the point mass m and the point

Talk:Free-fall time

Mistake in the derivation and expression of the equation for free-fall time

Latest comment: 6 years ago2 comments2 people in discussion

I suspect that in the following equation[1] someone messed up with the relation between orbital period, and the distance between the point mass m and the point source M.

1. t(ff) = (π/2) · R^(3/2) / ( 2G · (M + m))^(1/2)

I think that the correct expression would rather be:

1'. t(ff) = (π/2) · R^(3/2) / (G · (M + m))^(1/2)

Example (for comparison):

3. R = a = length of the orbit semi-major axis = 6371.0 km 4. μ = G·M = 398600.4418 km^3s^−2

M >> m

1'. t(ff) ≈ (3.14159/2) · R^(3/2) / μ^(1/2)

t(ff) ≈ (3.14159/2) · (6371.0 km)^1.5 / (398600.4418 km^3s^−2)^0.5

t(ff) ≈ (3.14159/2)*(6371.0^1.5)/((398600.4418)^0.5) = 1265.20829316 s

4 · t(ff) = 5060.83317264 s

Comparison with alternative equation[2]

2. T = 2π · (a^3 / μ)^(1/2)

T = 2 · 3.14159 · ( (6371.0 km)^3 / 398600.4418 km^3s−2 )^(1/2)

T = (2*3.14159*(6371.0^(3/2)))/(398600.4418^(1/2)) = 5060.83317263 s

References: 1. http://en.wikipedia.org/wiki/Free-fall_time
2. http://en.wikipedia.org/wiki/Semi-major_axis#Orbital_period
3. http://en.wikipedia.org/wiki/Earth (Earth's mean radius)
4. http://en.wikipedia.org/wiki/Standard_gravitational_parameter
5. http://en.wikipedia.org/wiki/Free_fall#Inverse-square_law_gravitational_field
37.133.53.224 (talk) 20:02, 19 January 2015 (UTC)Reply

It is not a mistake. The R in the article is the radial distance of the infalling particle, The R that you refer to is the semi-major axis, so R = 2R. The formulae in the article are then correct. FredV (talk) 12:00, 28 June 2020 (UTC)Reply

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