Need help with keplers 3rd law of periods problem

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In summary, the problem involves finding the radius of a planet with an unknown mass, given the period and radius of a satellite in orbit around it, and the magnitude of the gravitational acceleration on the planet's surface. Using Kepler's third law and the equation for gravitational acceleration, the mass of the planet can be solved for and then used to calculate the radius of the planet.
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[SOLVED] need help with keplers 3rd law of periods problem

Homework Statement


A 20 kg satellite has a circular orbit with a period of 2.6 h and a radius of 9.4E6 m around a planet of unknown mass. If the magnitude of the gravitational acceleration on the surface of the planet is 8.6 m/s2, what is the radius of the planet?



Homework Equations


T^2=((4pi^2)/(GM))r^3

a gravity=(GM)/r^2


The Attempt at a Solution



I used the law of periods to find the mass and then used the second equation to find the radius of the planet but I got a really small number
 
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  • #2
Can you show your calculations in more detail? Note that r in your first equation is not the same as r in the second (in case you forgot that).
 
  • #3
and I'm not sure if it's correct.

Hello,

It looks like you are on the right track with using Kepler's third law to solve this problem. Remember that in this equation, T represents the orbital period, M represents the mass of the central body (in this case, the planet), and r represents the distance between the satellite and the center of the planet.

To find the mass of the planet, you can rearrange the equation to solve for M: M = (4π^2r^3)/(GT^2). Plugging in the given values, we get M = (4π^2 * (9.4E6)^3)/(6.67E-11 * (2.6*3600)^2) = 3.1E24 kg.

Then, to find the radius of the planet, we can use the equation for gravitational acceleration: a = (GM)/r^2. Rearranging for r, we get r = √(GM/a). Plugging in the values, we get r = √((6.67E-11 * 3.1E24)/(8.6)) = 6.2E6 m. This is a reasonable answer, as it is smaller than the orbital radius of the satellite.

I hope this helps! Let me know if you have any further questions.
 

1. What is Kepler's 3rd law of periods?

Kepler's 3rd law of periods states that the square of the period of an orbiting object is directly proportional to the cube of its semi-major axis. In simpler terms, it relates the time it takes for an object to orbit around another object to the distance between them.

2. How do I calculate the period of an orbiting object using Kepler's 3rd law?

To calculate the period of an orbiting object, you will need to know the semi-major axis of the orbit (represented by a in the formula), which is the distance from the center of the orbit to the furthest point. Then, you can use the formula T^2/a^3 = 4π^2/GM, where T is the period, G is the gravitational constant, and M is the mass of the central object. Rearrange the formula to solve for T and you will have the period of the orbiting object.

3. Can Kepler's 3rd law be applied to all objects in orbit?

Yes, Kepler's 3rd law can be applied to any object in orbit around another object, regardless of their size or mass. This law is a fundamental principle in understanding the motion of objects in space.

4. What units should I use when applying Kepler's 3rd law?

When using Kepler's 3rd law, it is important to use consistent units. The period should be in seconds, the semi-major axis in meters, and the mass in kilograms. This will ensure that the equation is balanced and the correct result is obtained.

5. Can Kepler's 3rd law be used to calculate the period of objects in elliptical orbits?

Yes, Kepler's 3rd law can be used to calculate the period of objects in both circular and elliptical orbits. However, the semi-major axis used in the formula will be different for an elliptical orbit, as it is the average distance from the central object, not the furthest point. In this case, the semi-major axis is represented by a in the formula, but it is equal to half of the major axis of the ellipse.

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