Question

a satellite orbits earth at an altitude of 300 km. find its orbital speed and period.

Explanation:

Step1: Recall relevant formulas

The orbital speed \( v \) of a satellite is given by \( v = \sqrt{\frac{GM}{r}} \), where \( G = 6.67\times 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2 \) (gravitational constant), \( M = 5.97\times 10^{24}\ \text{kg} \) (mass of Earth), and \( r \) is the distance from the center of Earth to the satellite. The radius of Earth \( R = 6.37\times 10^{6}\ \text{m} \), so \( r=R + h \), where \( h = 300\times 10^{3}\ \text{m} \). The orbital period \( T \) is given by \( T=\frac{2\pi r}{v} \) or \( T = 2\pi\sqrt{\frac{r^3}{GM}} \).

Step2: Calculate \( r \)

\( r=R + h=6.37\times 10^{6}\ \text{m}+300\times 10^{3}\ \text{m}=6.67\times 10^{6}\ \text{m} \)

Step3: Calculate orbital speed \( v \)

Substitute \( G \), \( M \), and \( r \) into the speed formula:
\( v=\sqrt{\frac{6.67\times 10^{-11}\times5.97\times 10^{24}}{6.67\times 10^{6}}} \)
First, calculate the numerator: \( 6.67\times 10^{-11}\times5.97\times 10^{24}=6.67\times5.97\times 10^{13}\approx39.82\times 10^{13}=3.982\times 10^{14} \)
Then divide by \( r \): \( \frac{3.982\times 10^{14}}{6.67\times 10^{6}}\approx5.97\times 10^{7} \)
Take the square root: \( v=\sqrt{5.97\times 10^{7}}\approx7727\ \text{m/s}\approx7.73\ \text{km/s} \)

Step4: Calculate orbital period \( T \)

Using \( T = 2\pi\sqrt{\frac{r^3}{GM}} \)
First, calculate \( r^3=(6.67\times 10^{6})^3\approx6.67^3\times 10^{18}\approx296.0\times 10^{18}=2.96\times 10^{20}\ \text{m}^3 \)
\( GM = 6.67\times 10^{-11}\times5.97\times 10^{24}\approx3.98\times 10^{14}\ \text{N}\cdot\text{m}^2/\text{kg} \) (same as numerator in speed calculation)
\( \frac{r^3}{GM}=\frac{2.96\times 10^{20}}{3.98\times 10^{14}}\approx7.44\times 10^{5}\ \text{s}^2 \)
\( \sqrt{7.44\times 10^{5}}\approx863\ \text{s} \)
\( T = 2\pi\times863\approx5420\ \text{s}\approx1.51\ \text{hours} \) (or using \( T=\frac{2\pi r}{v} \), \( 2\pi\times6.67\times 10^{6}\approx4.19\times 10^{7}\ \text{m} \), divide by \( v = 7727\ \text{m/s} \), \( \frac{4.19\times 10^{7}}{7727}\approx5420\ \text{s} \), same result)

Answer:

Orbital speed: approximately \( 7.73\ \text{km/s} \) (or \( 7730\ \text{m/s} \)); Orbital period: approximately \( 5420\ \text{s} \) (or \( 1.51\ \text{hours} \))