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

Explanation:

Step1: Recall orbital speed formula

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\ \text{km}=300\times 10^{3}\ \text{m} ). Thus, ( r = 6.37\times 10^{6}+ 300\times 10^{3}=6.67\times 10^{6}\ \text{m} ).

Step2: Calculate orbital speed

Substitute ( G ), ( M ), and ( r ) into the formula: [ \begin{align*} v&=\sqrt{\frac{6.67\times 10^{-11}\times5.97\times 10^{24}}{6.67\times 10^{6}}}\ &=\sqrt{\frac{6.67\times5.97\times 10^{13}}{6.67\times 10^{6}}}\ &=\sqrt{9.0\times 10^{6}}\ &\approx 3.0\times 10^{3}\ \text{m/s} \ (or\ 3.0\ \text{km/s}) \end{align*} ]

Step3: Recall orbital period formula

The orbital period ( T ) is given by ( T=\frac{2\pi r}{v} ), or using ( T = 2\pi\sqrt{\frac{r^{3}}{GM}} ). Using the first formula with ( r = 6.67\times 10^{6}\ \text{m} ) and ( v\approx 3.0\times 10^{3}\ \text{m/s} ): [ \begin{align*} T&=\frac{2\pi\times6.67\times 10^{6}}{3.0\times 10^{3}}\ &\approx\frac{4.19\times 10^{7}}{3.0\times 10^{3}}\ &\approx 1.4\times 10^{4}\ \text{s} \end{align*} ] Converting to hours: ( T=\frac{1.4\times 10^{4}}{3600}\approx 3.9\ \text{hours} ), or using the second formula: [ \begin{align*} T&=2\pi\sqrt{\frac{(6.67\times 10^{6})^{3}}{6.67\times 10^{-11}\times5.97\times 10^{24}}}\ &=2\pi\sqrt{\frac{6.67^{3}\times 10^{18}}{6.67\times5.97\times 10^{13}}}\ &=2\pi\sqrt{\frac{6.67^{2}\times 10^{5}}{5.97}}\ &\approx 2\pi\sqrt{\frac{44.49\times 10^{5}}{5.97}}\ &\approx 2\pi\sqrt{7.45\times 10^{5}}\ &\approx 2\pi\times 863\ &\approx 5420\ \text{s}\ (approx\ 1.5\ \text{hours? Wait, no, earlier calculation had an error. Wait, correct calculation: } r = 6.37e6 + 0.3e6 = 6.67e6\ \text{m. } GM = 6.67e - 11\times5.97e24 = 3.98e14\ \text{m}^3/\text{s}^2. Then } r^3=(6.67e6)^3\approx 2.96e20\ \text{m}^3. Then } T = 2\pi\sqrt{\frac{2.96e20}{3.98e14}}=2\pi\sqrt{7.43e5}=2\pi\times 862\approx 5410\ \text{s}\approx 1.5\ \text{hours? Wait, no, the correct orbital speed for low Earth orbit (around 300 km) is about 7.7 km/s, and period about 90 minutes. Oh, I see, my mistake in Step2: I miscalculated ( v ). Let's recalculate ( v ):

( v=\sqrt{\frac{GM}{r}}=\sqrt{\frac{6.67e - 11\times5.97e24}{6.67e6}}=\sqrt{\frac{5.97e24}{6.67e6}\times6.67e - 11}=\sqrt{5.97e24\times1e - 17}=\sqrt{5.97e7}\approx 7.73e3\ \text{m/s}\ (7.73\ \text{km/s}) ). Then ( T=\frac{2\pi r}{v}=\frac{2\pi\times6.67e6}{7.73e3}\approx\frac{4.19e7}{7.73e3}\approx 5420\ \text{s}\approx 90\ \text{minutes}\ (1.5\ \text{hours}) ). So my initial Step2 had a calculation error. Let's redo Step2:

Step2 (corrected): Calculate orbital speed

( r = R + h = 6.37\times 10^{6}\ \text{m}+ 300\times 10^{3}\ \text{m}=6.67\times 10^{6}\ \text{m} ) ( GM = 6.67\times 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2\times5.97\times 10^{24}\ \text{kg}=3.98\times 10^{14}\ \text{m}^3/\text{s}^2 ) ( v=\sqrt{\frac{GM}{r}}=\sqrt{\frac{3.98\times 10^{14}}{6.67\times 10^{6}}}=\sqrt{5.97\times 10^{7}}\approx 7.73\times 10^{3}\ \text{m/s}\ (7.73\ \text{km/s}) )

Step3 (corrected): Calculate orbital period

( T=\frac{2\pi r}{v}=\frac{2\pi\times6.67\times 10^{6}}{7.73\times 10^{3}}\approx\frac{4.19\times 10^{7}}{7.73\times 10^{3}}\approx 5420\ \text{s}\approx 90\ \text{minutes}\ (or\ 1.5\ \text{hours}) )

Answer:

Orbital speed: approximately ( \boldsymbol{7.7\ \text{km/s}} ) (or ( 7.7\times 10^{3}\ \text{m/s} )), Orbital period: approximately ( \boldsymbol{90\ \text{minutes}} ) (or ( 5.4\times 10^{3}\ \text{s} ))