Question

a satellite has an orbital radius of 7.0×10⁶ m. calculate its orbital speed.

Explanation:

Step1: Recall the formula for orbital speed

For a satellite in circular orbit, the orbital speed \( v \) is given by \( v = \sqrt{\frac{GM}{r}} \), where \( G \) is the gravitational constant (\( G = 6.67\times 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2 \)), \( M \) is the mass of the central body (assuming Earth, \( M = 5.97\times 10^{24}\ \text{kg} \)), and \( r \) is the orbital radius.

Step2: Substitute the values into the formula

Given \( r = 7.0\times 10^{6}\ \text{m} \), \( G = 6.67\times 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2 \), and \( M = 5.97\times 10^{24}\ \text{kg} \).

First, calculate \( GM \):
\( GM=(6.67\times 10^{-11})\times(5.97\times 10^{24}) \)
\( GM\approx 6.67\times5.97\times 10^{-11 + 24} \)
\( GM\approx 39.82\times 10^{13}=3.982\times 10^{14}\ \text{m}^3/\text{s}^2 \)

Then, calculate \( \frac{GM}{r} \):
\( \frac{3.982\times 10^{14}}{7.0\times 10^{6}}=\frac{3.982}{7.0}\times 10^{14 - 6}\approx 0.569\times 10^{8}=5.69\times 10^{7}\ \text{m}^2/\text{s}^2 \)

Finally, take the square root:
\( v = \sqrt{5.69\times 10^{7}} \)
\( v\approx\sqrt{5.69}\times 10^{3.5} \) (since \( \sqrt{10^{7}} = 10^{3.5} \))
\( \sqrt{5.69}\approx 2.385 \), \( 10^{3.5}=10^{3}\times\sqrt{10}\approx 3162.3 \)
\( v\approx 2.385\times 3162.3\approx 7540\ \text{m/s} \) (or we can also use the approximation for Earth - orbiting satellites, the orbital speed formula can also be approximated as \( v=\sqrt{\frac{gR^{2}}{r}} \), where \( g = 9.8\ \text{m/s}^2 \), \( R = 6.4\times 10^{6}\ \text{m} \). Let's check this way:

\( gR^{2}=(9.8)\times(6.4\times 10^{6})^{2}=9.8\times 40.96\times 10^{12}=401.408\times 10^{12}=4.01408\times 10^{14}\ \text{m}^3/\text{s}^2 \)

\( \frac{gR^{2}}{r}=\frac{4.01408\times 10^{14}}{7.0\times 10^{6}}\approx 5.734\times 10^{7}\ \text{m}^2/\text{s}^2 \)

\( v = \sqrt{5.734\times 10^{7}}\approx 7572\ \text{m/s} \), the slight difference is due to the approximation of \( G \) and \( M \) values. The more accurate value using \( G \) and \( M \) is around \( 7.5\times 10^{3}\ \text{m/s} \) (or \( 7.5\ \text{km/s} \))

Answer:

The orbital speed of the satellite is approximately \(\boldsymbol{7.5\times 10^{3}\ \text{m/s}}\) (or \(7500\ \text{m/s}\))