Chapter 8, Problem 79RE

Solution

To find: the velocity of the satellite at its apogee and perigee.

The velocity of the satellite at its apogee and perigee $2634$ km/sec, at the apogee is 9788 m/sec.

Given information:

The radius of Earth is 6380 km.

The maximum altitude of Earth satellite (at apogee) is 18,000 km.

The minimum altitude of Earth satellite (at perigee) is 170 km.

Formula used:

• The velocity of a body in an elliptical Earth orbit at a distance $r$ (in meters) from the focus (the center of Earth) is
$v=\sqrt{3.99\times {10}^{14}\left(\frac{2}{r}-\frac{1}{a}\right)}$ m/sec,

where $a$ is the semimajor axis of the ellipse.

Calculation:

The major axis of Earth satellite is $18,000+2\left(6380\right)+170=30930$ . Thus, semimajor axis $a$ of the Earth satellite is 15465 m.

  

The velocity of the satellite at its apogee:

The distance of the satellite at its apogee from the center of Earth is 18,000+6380 = 24380 km, which is equal to $24380\times {10}^3$ m.

Substitute the values $24380\times {10}^3$ for $r$ , $15465\times {10}^3$ for $a$ in the equation $v=\sqrt{3.99\times {10}^{14}\left(\frac{2}{r}-\frac{1}{a}\right)}$ .

Thus,

  $\begin{array}{c}v=\sqrt{3.99\times {10}^{14}\left(\frac{2}{r}-\frac{1}{a}\right)}\\ =\sqrt{3.99\times {10}^{14}\left(\frac{2}{24380\times {10}^3}-\frac{1}{15465\times {10}^3}\right)}\\ =\sqrt{3.99\times {10}^{11}\left(0.1737\right)}\\ =\sqrt{0.693063\times {10}^{11}}\\ =2.6326\times {10}^5\\ \approx 2634\end{array}$

Hence, the velocity of the satellite at the apogee is $2634$ km/sec.

The velocity of the satellite at its perigee:

The distance of satellite at its perigee from the center of the Earth is 6380+170 = 6550 km.

Substitute the values $6550\times {10}^3$ for $r$ , for $a$ in the equation $v=\sqrt{3.99\times {10}^{14}\left(\frac{2}{r}-\frac{1}{a}\right)}$ .

Thus,

  $\begin{array}{c}v=\sqrt{3.99\times {10}^{14}\left(\frac{2}{r}-\frac{1}{a}\right)}\\ =\sqrt{3.99\times {10}^{14}\left(\frac{2}{6550\times {10}^3}-\frac{1}{15465\times {10}^3}\right)}\\ =\sqrt{3.99\times {10}^{11}\left(2.4\times {10}^{-4}\right)}\\ =\sqrt{95.76\times {10}^6}\\ =9.788\times {10}^3\\ =9788\end{array}$

Hence, the velocity of the satellite at the apogee is 9788 m/sec.