Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 7 221 - 252 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 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}} ## 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.