thatgirlyouknow said:Homework Statement
A satellite is in a circular orbit about the Earth (M = 5.98 x 1024 kg). The period of the satellite is 9.93 x 104 s. What is the speed at which the satellite travels?
Homework Equations
v=(2*pi*r)/T
v=sqrt((GMe)/r)
The Attempt at a Solution
sqrt[((6.67x10^-11)*(5.98x10^24))/r) = 2*pi*r / T
T = 2piR^3/2 / (sqrt[GMe])
r = 8.414 x 10^15
thatgirlyouknow said:Ok I worked it again and got v = 2933.27 m/s. (Hooray human error). Can anyone verify that as correct?
The speed of a satellite in circular orbit is the rate at which the satellite travels around the Earth in a perfect circular path.
Yes, the speed of a satellite in circular orbit varies depending on the altitude of the orbit. Satellites in higher altitudes have a slower speed, while satellites in lower altitudes have a faster speed.
The speed of a satellite in circular orbit can be calculated using the formula v = √(G*M/R), where G is the gravitational constant, M is the mass of the Earth, and R is the radius of the orbit.
The speed of a satellite in circular orbit is affected by the mass of the Earth, the altitude of the orbit, and the gravitational constant. Additionally, the shape and size of the Earth and any external forces, such as atmospheric drag, can also impact the speed.
The speed of a satellite in circular orbit is directly proportional to its orbital period. This means that as the speed increases, the orbital period also increases. Conversely, if the speed decreases, the orbital period also decreases.