Transit satellite navigation

Problems in Exploration Seismology and their Solutions

Series	Geophysical References Series
Title	Problems in Exploration Seismology and their Solutions
Author	Lloyd P. Geldart and Robert E. Sheriff
Chapter	7
Pages	221 - 252
DOI	http://dx.doi.org/10.1190/1.9781560801733
ISBN	ISBN 9781560801153
Store	SEG Online Store

Problem 7.3a

Determine the acceleration of gravity at the orbit of a Transit satellite 1070 km above the Earth, knowing that ${\textit {g}}$ at the surface of the Earth is 9.81 m/s$^{2}$, and that the gravitational force varies inversely as the square of the distance between the centers of gravity of the masses. The radius of the Earth is 6370 km.

Background

A satellite is in a stable orbit around the Earth when the gravitational force $({\textit {mg}})$ pulling it earthward equals the centrifugal force $mV^{2}/R$, where $g$ is the acceleration of gravity, $m$ and $V$ the satellite’s mass and velocity, and $R$ the radius of its orbit about the center of the Earth.

Solution

The radius of the satellite’s orbit is $\left(6370+1070\right)=7440$ km. Since $g$ is proportional to the force of gravity, at the satellite’s orbit,

${\begin{aligned}g=9.81(6370/7440)^{2}=7.19\ \mathrm {m/s} ^{2}.\end{aligned}}$

Problem 7.3b

What is the satellite’s velocity if its orbit is stable?

Solution

For a stable orbit, the gravitational acceleration is balanced by the centripetal acceleration.

Thus,

${\begin{aligned}V^{2}/R=g=7.19\ \mathrm {m/s} ^{2},\\V=(7.19\times 10^{-3}\times 7440)^{1/2}=7.31\ \mathrm {km/s} .\end{aligned}}$

Problem 7.3c

How long does it take for one orbit?

Solution

The length of the nearly circular orbit is $2\pi \times 7440$ km, so the time for one orbit is

${\begin{aligned}T=2\pi \times 7440/7.31=6395\ \mathrm {s} =106\ \mathrm {minutes} ,35\ \mathrm {seconds} .\end{aligned}}$

Problem 7.3d

How far away is the satellite when it first emerges over the horizon?

Solution

In Figure 7.3a, ${\textit {O}}$ is the point of observation. The satellite first becomes visible when it reaches the tangent to the Earth at ${\textit {O}}$. The tangent is normal to the radius at ${\textit {O}}$, so

${\begin{aligned}x=(7440^{2}-6370^{2})^{1/2}=3840\,{\rm {km.}}\end{aligned}}$

Figure 7.3a  First visibility of satellite.

Problem 7.3e

What is the maximum time of visibility on a single satellite pass?

Solution

In Figure 7.3a the angle subtended at the center of the Earth by ${\textit {x}}$ is

${\begin{aligned}\theta =\cos ^{-1}\left(6370/7440\right)=31.1^{\circ }.\end{aligned}}$

The satellite is visible while it traverses an arc subtending $2\theta =62.2^{\circ }$. Since the entire orbit is traversed in 6395 s, the time of visibility is $6395\left(62.2/360\right)=1105\ \mathrm {s} =18$ minutes, 25 seconds.

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Characteristics of seismic events	Reflection field methods
Table of Contents (book)

Also in this chapter

• Radiolocation errors because of velocity variations
• Effect of station angle on location errors
• Transit satellite navigation
• Effective penetration of profiler sources
• Directivity of linear sources
• Sosie method
• Energy from an air-gun array
• Dominant frequencies of marine sources
• Effect of coil inductance on geophone equation
• Streamer feathering due to cross-currents
• Filtering effect of geophones and amplifiers
• Filter effects on waveshape
• Effect of filtering on event picking
• Binary numbers

External links

find literature about Transit satellite navigation