Physics Derivation Graph navigation Sign in

radius for satellite in geostationary orbit

Generated by the Physics Derivation Graph.

https://en.wikipedia.org/wiki/Geostationary_orbit#Derivation_of_geostationary_altitude

Change of variable \(F\) to \(F_{\rm gravity}\) and \(m_1\) to \(m_{\rm Earth}\) and \(m_2\) to \(m_{\rm satellite}\) and \(x\) to \(r\) in Eq. \ref{eq:2820438}; yields Eq. \ref{eq:F0}. \begin{equation} F_{\rm gravity} = G \frac{m_{\rm Earth} m_{\rm satellite}}{r^2} \label{eq:1917654} \end{equation} Eq. \ref{eq:8242154} is an initial equation. \begin{equation} F = \frac{m v^2}{r} \label{eq:8242154} \end{equation} Change of variable \(F\) to \(F_{\rm centripetal}\) and \(m\) to \(m_{\rm satellite}\) and \(v\) to \(v_{\rm satellite}\) and \(8242154\) to \(6845877\) in Eq. \ref{eq:#9}; yields Eq. \ref{eq:F0}. \begin{equation} F_{\rm centripetal} = \frac{m_{\rm satellite} v_{\rm satellite}^2}{r} \label{eq:6845877} \end{equation} Eq. \ref{eq:2154616} is an assumption. \begin{equation} F_{\rm centripetal} = F_{\rm gravity} \label{eq:2154616} \end{equation} Substitute LHS of Eq. \ref{eq:1917654} and LHS of Eq. \ref{eq:6845877} into Eq. \ref{eq:2154616}; yields Eq. \ref{eq:4948724}. \begin{equation} \frac{m_{\rm satellite} v_{\rm satellite}^2}{r} = G \frac{m_{\rm Earth} m_{\rm satellite}}{r^2} \label{eq:4948724} \end{equation} Change variable \(C\) to \(d\) in Eq. \ref{eq:1115424}; yields Eq. \ref{eq:5369477}. \begin{equation} d = 2 \pi r \label{eq:5369477} \end{equation} Substitute LHS of Eq. \ref{eq:5369477} into Eq. \ref{eq:5114041}; yields Eq. \ref{eq:8090893}. \begin{equation} v = \frac{2 \pi r}{t} \label{eq:8090893} \end{equation} Change variable \(t\) to \(T_{\rm orbit}\) in Eq. \ref{eq:8090893}; yields Eq. \ref{eq:2392562}. \begin{equation} v = \frac{2 \pi r}{T_{\rm orbit}} \label{eq:2392562} \end{equation} Raise both sides of Eq. \ref{eq:2392562} to \(2\); yields Eq. \ref{eq:6390693}. \begin{equation} v^2 = \frac{4 \pi^2 r^2}{T_{\rm orbit}^2} \label{eq:6390693} \end{equation} Divide both sides of Eq. \ref{eq:4948724} by \(\frac{m_{\rm satellite}}{r}\); yields Eq. \ref{eq:2009493}. \begin{equation} v_{\rm satellite}^2 = G \frac{m_{\rm Earth}}{r} \label{eq:2009493} \end{equation} LHS of Eq. \ref{eq:2009493} is equal to LHS of Eq. \ref{eq:6390693}; yields Eq. \ref{eq:2871066}. \begin{equation} G \frac{m_{\rm Earth}}{r} = \frac{4 \pi^2 r^2}{T_{\rm orbit}^2} \label{eq:2871066} \end{equation} Multiply both sides of Eq. \ref{eq:2871066} by \(r T_{\rm orbit}^2\); yields Eq. \ref{eq:7188516}. \begin{equation} T_{\rm orbit}^2 G m_{\rm Earth} = 4 \pi^2 r^3 \label{eq:7188516} \end{equation} Divide both sides of Eq. \ref{eq:7188516} by \(4 \pi^2\); yields Eq. \ref{eq:6238570}. \begin{equation} \frac{T_{\rm orbit}^2 G m_{\rm Earth}}{4 \pi^2} = r^3 \label{eq:6238570} \end{equation} Raise both sides of Eq. \ref{eq:6238570} to \(1/3\); yields Eq. \ref{eq:7139326}. \begin{equation} \left(\frac{T_{\rm orbit}^2 G m_{\rm Earth}}{4 \pi^2}\right)^{1/3} = r \label{eq:7139326} \end{equation} Eq. \ref{eq:9978909} is an assumption. \begin{equation} T_{\rm geostationary orbit} = 24\ {\rm hours} \label{eq:9978909} \end{equation} Change variable \(T_{\rm orbit}\) to \(T_{\rm geostationary\ orbit}\) and \(r\) to \(r_{\rm geostationary\ orbit}\) in Eq. \ref{eq:7139326}; yields Eq. \ref{eq:4507350}. \begin{equation} \left(\frac{T_{\rm geostationary\ orbit}^2 G m_{\rm Earth}}{4 \pi^2}\right)^{1/3} = r_{\rm geostationary\ orbit} \label{eq:4507350} \end{equation}