Question
Find the time of one revolution for an earth satellite moving in a circular orbit with an orbital speed of 7500 m/s. If the answer is not available, ask a new question with the given orbital speed and request an answer
Answers
GPT-4o mini
To find the time of one revolution (the orbital period) for a satellite moving in a circular orbit, we can use the formula:
\[
T = \frac{2\pi r}{v}
\]
where:
- \( T \) is the orbital period,
- \( r \) is the radius of the orbit, and
- \( v \) is the orbital speed.
However, we don't have the radius of the orbit directly. To find the radius, we can use the gravitational force that provides the necessary centripetal force for circular motion, using the following formula for the orbital speed:
\[
v = \sqrt{\frac{GM}{r}}
\]
where:
- \( G \) is the gravitational constant (\(6.674 \times 10^{-11} \, \text{m}^3/\text{kg}\cdot\text{s}^2\)),
- \( M \) is the mass of the Earth (approximately \(5.972 \times 10^{24} \, \text{kg}\)).
First, we can rearrange this formula to solve for \( r \):
\[
r = \frac{GM}{v^2}
\]
Now let's calculate \( r \) using the given orbital speed \( v = 7500 \, \text{m/s} \).
Substituting the values:
\[
r = \frac{(6.674 \times 10^{-11} \, \text{m}^3/\text{kg}\cdot\text{s}^2)(5.972 \times 10^{24} \, \text{kg})}{(7500 \, \text{m/s})^2}
\]
Calculating this gives:
\[
r \approx \frac{3.986 \times 10^{14} \, \text{m}^3/\text{s}^2}{56250000 \, \text{m}^2/\text{s}^2} \approx 7075 \, \text{km}
\]
Now substituting \( r \) back into the formula for the orbital period \( T \):
\[
T = \frac{2\pi r}{v} = \frac{2\pi(7075 \times 10^3 \, \text{m})}{7500 \, \text{m/s}}
\]
Calculating \( T \):
\[
T \approx \frac{2\pi(7075000)}{7500} \approx \frac{44413.25 \, \text{m}}{7500} \approx 5921.77 \, \text{s}
\]
To convert seconds to hours:
\[
T \approx \frac{5921.77 \, \text{s}}{3600} \approx 1.64 \, \text{hours}
\]
So, the time of one revolution for the Earth satellite moving at an orbital speed of 7500 m/s is approximately **1.64 hours**.
\[
T = \frac{2\pi r}{v}
\]
where:
- \( T \) is the orbital period,
- \( r \) is the radius of the orbit, and
- \( v \) is the orbital speed.
However, we don't have the radius of the orbit directly. To find the radius, we can use the gravitational force that provides the necessary centripetal force for circular motion, using the following formula for the orbital speed:
\[
v = \sqrt{\frac{GM}{r}}
\]
where:
- \( G \) is the gravitational constant (\(6.674 \times 10^{-11} \, \text{m}^3/\text{kg}\cdot\text{s}^2\)),
- \( M \) is the mass of the Earth (approximately \(5.972 \times 10^{24} \, \text{kg}\)).
First, we can rearrange this formula to solve for \( r \):
\[
r = \frac{GM}{v^2}
\]
Now let's calculate \( r \) using the given orbital speed \( v = 7500 \, \text{m/s} \).
Substituting the values:
\[
r = \frac{(6.674 \times 10^{-11} \, \text{m}^3/\text{kg}\cdot\text{s}^2)(5.972 \times 10^{24} \, \text{kg})}{(7500 \, \text{m/s})^2}
\]
Calculating this gives:
\[
r \approx \frac{3.986 \times 10^{14} \, \text{m}^3/\text{s}^2}{56250000 \, \text{m}^2/\text{s}^2} \approx 7075 \, \text{km}
\]
Now substituting \( r \) back into the formula for the orbital period \( T \):
\[
T = \frac{2\pi r}{v} = \frac{2\pi(7075 \times 10^3 \, \text{m})}{7500 \, \text{m/s}}
\]
Calculating \( T \):
\[
T \approx \frac{2\pi(7075000)}{7500} \approx \frac{44413.25 \, \text{m}}{7500} \approx 5921.77 \, \text{s}
\]
To convert seconds to hours:
\[
T \approx \frac{5921.77 \, \text{s}}{3600} \approx 1.64 \, \text{hours}
\]
So, the time of one revolution for the Earth satellite moving at an orbital speed of 7500 m/s is approximately **1.64 hours**.