Convert Foot per second to Escape velocity (Earth)
Unit definitions
Foot per second
A foot is a unit of length equal to $0.304\,8 \ \text{m}$. Foot per second is the distance traveled in feet in one second.
Escape velocity (Earth)
The escape velocity is the minimum speed required to overcome the effects of gravity from the surface without falling back. The escape velocity of Earth is equal to $11.186\ \text{km/s}$.
How to convert Foot per second to Escape velocity (Earth)
$$\text{Speed}_{\text{Escape Velocityₑₐᵣₜₕ}} = \text{Speed}_{\text{fts⁻¹}} \cdot 0.00002724834614696942606829966028964777$$
Examples
Example 1
Convert $35.0\ \text{fts⁻¹}$ to $\text{Escape Velocityₑₐᵣₜₕ}$.
$$\text{Speed}_{\text{Escape Velocityₑₐᵣₜₕ}} = 35.0 \cdot 0.00002724834614696942606829966028964777$$
$$\text{Speed}_{\text{Escape Velocityₑₐᵣₜₕ}} = 0.000954$$
$$\therefore \ 35.0\ \text{fts⁻¹} = 0.000954 \ \text{Escape Velocityₑₐᵣₜₕ}$$
Example 2
Convert $85.0\ \text{fts⁻¹}$ to $\text{Escape Velocityₑₐᵣₜₕ}$.
$$\text{Speed}_{\text{Escape Velocityₑₐᵣₜₕ}} = 85.0 \cdot 0.00002724834614696942606829966028964777$$
$$\text{Speed}_{\text{Escape Velocityₑₐᵣₜₕ}} = 0.002316$$
$$\therefore \ 85.0\ \text{fts⁻¹} = 0.002316 \ \text{Escape Velocityₑₐᵣₜₕ}$$
Example 3
Convert $105.0\ \text{fts⁻¹}$ to $\text{Escape Velocityₑₐᵣₜₕ}$.
$$\text{Speed}_{\text{Escape Velocityₑₐᵣₜₕ}} = 105.0 \cdot 0.00002724834614696942606829966028964777$$
$$\text{Speed}_{\text{Escape Velocityₑₐᵣₜₕ}} = 0.002861$$
$$\therefore \ 105.0\ \text{fts⁻¹} = 0.002861 \ \text{Escape Velocityₑₐᵣₜₕ}$$
Example 4
Convert $135.0\ \text{fts⁻¹}$ to $\text{Escape Velocityₑₐᵣₜₕ}$.
$$\text{Speed}_{\text{Escape Velocityₑₐᵣₜₕ}} = 135.0 \cdot 0.00002724834614696942606829966028964777$$
$$\text{Speed}_{\text{Escape Velocityₑₐᵣₜₕ}} = 0.003679$$
$$\therefore \ 135.0\ \text{fts⁻¹} = 0.003679 \ \text{Escape Velocityₑₐᵣₜₕ}$$
Example 5
Convert $170.0\ \text{fts⁻¹}$ to $\text{Escape Velocityₑₐᵣₜₕ}$.
$$\text{Speed}_{\text{Escape Velocityₑₐᵣₜₕ}} = 170.0 \cdot 0.00002724834614696942606829966028964777$$
$$\text{Speed}_{\text{Escape Velocityₑₐᵣₜₕ}} = 0.004632$$
$$\therefore \ 170.0\ \text{fts⁻¹} = 0.004632 \ \text{Escape Velocityₑₐᵣₜₕ}$$