Calculating the Orbital Period

In summary, To calculate the Orbital Period for Earth's orbit around the Sun using Newton's second law and the Universal Law of Gravitation, you need to set the Centrifugal Force and the Force of Gravity equal to each other. After cancelling out the mass of the Earth, you can rearrange the equation to solve for the Orbital Period. Once you have the equation v= 2πr/t, you can substitute it into the equation and solve for the Orbital Period.
  • #1
Sci21
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Homework Statement


Calculate the Orbital Period for the following: Earth's orbit to the Sun. I get stuck towards the end. I must use Newton's second law and have the Universal Law of Gravitation equal the Centrifugal Force.
Newton's gravitational constant: G= 6.67*10^-11 Nm^2/kg^2
Mass of Sun = 1.98*10^30 kg
Mass of the Earth 5.97*10^24 kg
Distance of the Earth from the Sun: 149.6*10^6
T = time


Homework Equations


Centrifugal Force = (mv^2)/r
v=(2(pi)(r))t
Force of Gravity = (GMm)/d^2
Mm being mass, and G/d^2 being acceleration.


The Attempt at a Solution


I multiplied Newton's gravitational constant by the mass of the Earth by the mass of the Sun, and then divided it all by the distance of the Earth from the Sun. I got 5.270280882*10^35 km. This was for the gravitational pull.

For the Centrifugal Force, I multiplied the mass of the Earth by v=((2(pi)(r))^2)/t^2. And put it over 149.6*10^6 km. I got (3.52229083*10^34 kg*km)/t^2.

I then set my two results equal to each other. Next, I multiplied by t^2 as that is the variable I am trying to find. This gave me 5.270280882*10^35 km*t^2 = 3.5229083*10^34 kg*km. Then, I divided by km on both sides to cancel it. I now have 5.270280882*10^35 t^2 = 3.5229083*10^34 kg. If I divide by the number on the left side of the equation and then square root both sides, I get a number than can't possibly be the Orbital Period of the Earth.


Can you tell where I am going wrong? Any advice is appreciated.
Thanks.
 
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  • #2
Centrifugal Force = (mv^2)/r
Force of Gravity = (GMm)/d^2

For an orbit these two are equal so you have
mv^2/r = GMm/r^2

The mass of the Earth cancels, orbit's don't depend on the mass of the small object - that's why a spaceman and a space shuttle can float along in the same orbit.

v^2/r = GM/r^2

Work out v in terms of the circumference and the period ( v= 2 PI r/t) , do a bit of rearranging and you're there.
 
  • #3
OK, thanks. I will try this and let you know how it goes.
 
  • #4
((4(pi^2)(149.6 km^2))/t^2)/149.6 km=(6.67*10^-11(1.98*10^30 kg))/149.6 km^2

I do not understand what to do. If I cancel the 149.6 km from the denom. on the left side, I can't go much further.
 

1. What is the formula for calculating the orbital period?

The formula for calculating the orbital period is: T = 2π * √(a^3/GM), where T is the orbital period, a is the semi-major axis, G is the gravitational constant, and M is the mass of the central body.

2. What is the unit of measurement for orbital period?

The unit of measurement for orbital period is time, typically in seconds, minutes, hours, or days.

3. How do you calculate the semi-major axis?

The semi-major axis can be calculated by taking the average distance between the object and the central body at the closest and farthest points of its orbit. This can be done using data from observations or by using Kepler's third law: a = (G * M * T^2)/(4π^2), where a is the semi-major axis, G is the gravitational constant, M is the mass of the central body, and T is the orbital period.

4. Can the orbital period of an object change?

Yes, the orbital period of an object can change due to factors such as gravitational interactions with other objects, changes in the mass of the central body, or changes in the semi-major axis of the orbit.

5. Why is calculating the orbital period important in astronomy?

Calculating the orbital period is important in astronomy because it allows us to predict the motion of objects in our solar system and beyond. It also helps us understand the relationship between an object's distance from its central body and its orbital speed, which can provide insights into the formation and evolution of planetary systems.

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