Orbital period questions continued

In summary, the question asks for the orbital period of a 9.0x10^3kg satellite with an orbital radius of 3.20x10^7m and an altitude of 2.56x10^7m. The necessary equations to solve for the period are Kepler's third law and the formula for centripetal acceleration. By using these equations and the given values, the orbital period can be calculated to be 15.79 hours.
  • #1
Inertialforce
68
2

Homework Statement


A 9.0x10^3kg satellite with an orbital radius of 3.20x10^7m orbits the Earth at an altitude of 2.56x10^7m. What is the orbital period?

Homework Equations


ΣFc = mac
Kepler's third law equation

The Attempt at a Solution


For this question I know that it is asking me to find the orbital period, but it has also given me an altitude and mass for the satellite. Are they there just to through me off because you are supposed to use Kepler's third law equation to find the orbital period?

Or are they important and I am supposed to use the equation:

Fg = 4(pie)^2r/T^2

Basically what I am wondering is which equation do we use?
 
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  • #2
What does Kepler's third law give you? In relation to what you are looking for? What are the inputs required to use Kepler's Law?

~Lyuokdea
 
  • #3
Keplers third law: "The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit." i don't think it is necessary to use keplars law to solve this question, this is how i would figure it out;

ok so you can solve how fast the satellite is moving in relation the Earth: v=(GMe/r)^.5
(G=6.67x10^-11 gravitational constant) so plugging the numbers in gets you a velocity of 3536.4176...ms^-1, next we use the formula Ac=4[tex]\pi[/tex]^2r/T^2 like you had before but you can solve the acceleration by using Ac=v^2/r now that you have the velocity. This gives you 0.39082m/s^2 now put this back into your original formula which gives you a period of 15.79 hours

hope this helps
 

1. What is the orbital period and how is it calculated?

The orbital period is the time it takes for an object to complete one full orbit around another object. It can be calculated using Kepler's Third Law, which states that the square of the orbital period is directly proportional to the cube of the semi-major axis.

2. What factors affect the orbital period of a planet or satellite?

The orbital period of a planet or satellite is primarily affected by two factors: the mass of the central object it is orbiting and the distance between the two objects. Other factors such as the shape of the orbit and the presence of other objects can also have a minor impact.

3. Can the orbital period of a planet or satellite change over time?

Yes, the orbital period of a planet or satellite can change over time due to various factors such as gravitational interactions with other objects, changes in the shape of the orbit, and tidal forces. However, these changes are usually very small and take a long time to become noticeable.

4. How does the orbital period affect the climate of a planet?

The orbital period of a planet can affect its climate in several ways. For example, a longer orbital period can result in more extreme seasonal changes, while a shorter orbital period can lead to a more stable climate. The tilt of a planet's axis also plays a role in determining its climate.

5. Can the orbital period of a planet or satellite be used to determine its distance from the central object?

Yes, the orbital period of a planet or satellite can be used to determine its distance from the central object using Kepler's Third Law. By knowing the mass of the central object, the orbital period, and the semi-major axis, the distance between the two objects can be calculated. This is known as Kepler's Third Law of Planetary Motion.

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