Explain what the terms centrifugal and centripetal mean with regard to a satellite in orbit around the earth.
A satellite is in a circular orbit around the earth. The altitude of the satellite’s orbit above the surface of the earth is 1,400 km. (i) What are the centripetal and centrifugal accelerations acting on the satellite in its orbit? Give your answer in \(m/s^2\). (ii) What is the velocity of the satellite in this orbit? Give your answer in km/s. (iii) What is the orbital period of the satellite in this orbit? Give your answer in hours, minutes, and seconds. Note: assume the average radius of the earth is 6,378.137 km and Kepler’s constant has the value \(3.986004418 × 105 km^3/s^2\).
A satellite is in a circular orbit around the earth. The altitude of the satellite’s orbit above the surface of the earth is 1,400 km. (i) What are the centripetal and centrifugal accelerations acting on the satellite in its orbit? Give your answer in \(m/s^2\). (ii) What is the velocity of the satellite in this orbit? Give your answer in km/s. (iii) What is the orbital period of the satellite in this orbit? Give your answer in hours, minutes, and seconds. Note: assume the average radius of the earth is 6,378.137 km and Kepler’s constant has the value \(3.986004418 × 105 km^3/s^2\).
Solution to question 1:
In the case of a satellite orbiting the earth, the centrifugal force on the satellite is a force on the satellite that is directly away from the center of gravity of the earth and the centripetal force is one directly towards the center of gravity of the earth. The centrifugal force on a satellite will therefore try to fling the satellite away from the earth while the centripetal force will try to bring the satellite down towards the earth.
(i) We know that centripetal acceleration \(a = μ/r^2\), where μ is Kepler's constant.
The value of \(r = 6,378.137 + 1,400 = 7,778.137 km\),
thus centripetal acceleration \(a = \frac{3.986004418 × 10^5}{(7,778.137)^2}\) $$= 0.0065885 km/s^2 = 6.5885007 m/s^2$$ the centrifugal acceleration is given by \(a = v^2/r\), where v = the velocity of the satellite in a circular orbit.
Now velocity \(v = (μ/r)^{1/2}\) \(= \left(\frac{3.986004418 × 10^5}{7,778.137}\right)^{1/2}\) \(= 7.1586494 km/s\)
and so centrifugal acceleration \(a = \frac{(7.1586494)^2}{7,778.137}\) $$= 0.0065885007 km/s^2 = 6.5885007 m/s^2$$ NOTE: since the satellite was in stable orbit, the centrifugal acceleration must be equal to the centripetal acceleration, which we have found to be true here (but we needed only to calculate one of them).
(ii) We have already found out the velocity of the satellite in orbit in part (i) to be 7.1586494 km/s.
(iii) the orbital period \(T = \left[\frac{2πr^3}{μ}\right]^{1/2}\) $$= \left[\frac{2π × (7,778.137)^3}{3.986004418 × 10^5}\right]^{1/2}$$ \(= 6,826.912916 s\) = 1 hours 53 minutes 46.92 seconds.
In the case of a satellite orbiting the earth, the centrifugal force on the satellite is a force on the satellite that is directly away from the center of gravity of the earth and the centripetal force is one directly towards the center of gravity of the earth. The centrifugal force on a satellite will therefore try to fling the satellite away from the earth while the centripetal force will try to bring the satellite down towards the earth.
(i) We know that centripetal acceleration \(a = μ/r^2\), where μ is Kepler's constant.
The value of \(r = 6,378.137 + 1,400 = 7,778.137 km\),
thus centripetal acceleration \(a = \frac{3.986004418 × 10^5}{(7,778.137)^2}\) $$= 0.0065885 km/s^2 = 6.5885007 m/s^2$$ the centrifugal acceleration is given by \(a = v^2/r\), where v = the velocity of the satellite in a circular orbit.
Now velocity \(v = (μ/r)^{1/2}\) \(= \left(\frac{3.986004418 × 10^5}{7,778.137}\right)^{1/2}\) \(= 7.1586494 km/s\)
and so centrifugal acceleration \(a = \frac{(7.1586494)^2}{7,778.137}\) $$= 0.0065885007 km/s^2 = 6.5885007 m/s^2$$ NOTE: since the satellite was in stable orbit, the centrifugal acceleration must be equal to the centripetal acceleration, which we have found to be true here (but we needed only to calculate one of them).
(iii) the orbital period \(T = \left[\frac{2πr^3}{μ}\right]^{1/2}\) $$= \left[\frac{2π × (7,778.137)^3}{3.986004418 × 10^5}\right]^{1/2}$$ \(= 6,826.912916 s\) = 1 hours 53 minutes 46.92 seconds.