Solve Kepler's 3rd Law for Satellite Weight

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In summary, the satellite has a mass of 6000 kg and is in a circular orbit 4.40 * 10^5 m above the surface of a planet with a radius of 4.10 * 10^6 m. Using Kepler's third law, the mass of the planet can be calculated to be 1.06 * 10^24 kg. Using this value, the weight of the satellite when it is at rest on the planet's surface can be calculated to be 2.523569 * 10^4 N or approximately 25235.69 N.
  • #1
ny_aish
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A satellite has a mass of 6000 kg and is in a circular orbit 4.40 10^5 m above the surface of a planet. The period of the orbit is two hours. The radius of the planet is 4.10 10^6 m. What is the true weight of the satellite when it is at rest on the planet's surface?

m=6000kg
t=7200 s

Mp = 4(3.14)^2 (4.40 * 10^5 + 4.10 * 10^ 6)^3/(6.67*10^-11)(7200)^2
Mp = 1.06 * 10^24
then using Mp for Wp (weight of planet)

Wp= (6.67*10^-11)(1.06 * 10^24)(6000) / (4.40 * 10^5)^2
* I used Keplers 3rd law to get the Mass of planet and then applied it to Weight of planet from which I got 2.191 x 10^6 and it still marks it wrong
 
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  • #2
You put in the height of the satellite instead of the radius of the planet.
 
  • #3
ohhhhhh
 
  • #4
so its (6.67*10^-11)(1.06*10^24)(6000) / (4.10*10^6)^2
which is = 2.523569 * 10^4 which is same as 25235.69 ... can you please just confirm it for me? I have only 1 chance left, and I dnt want to lose points. Thanks
 
  • #5
Seems good to me.
 
  • #6
thanx
 

1. What is Kepler's 3rd Law for Satellite Weight?

Kepler's 3rd Law, also known as the Law of Periods, states that the square of a satellite's orbital period is proportional to the cube of its average distance from the object it is orbiting, such as a planet or star. This law is expressed mathematically as: T^2 = k * r^3, where T is the orbital period in seconds, r is the average distance in meters, and k is a constant.

2. How can Kepler's 3rd Law be used to solve for a satellite's weight?

By rearranging the equation T^2 = k * r^3, we can solve for the constant k, which is dependent on the mass of the object being orbited. Then, using the known value of the constant k and the satellite's orbital period and average distance, we can solve for the satellite's weight using the equation W = k * (r/T)^3, where W is the weight in kilograms.

3. What are the units of measurement for the variables in Kepler's 3rd Law?

The units of measurement for the orbital period T are in seconds, the average distance r is in meters, and the weight W is in kilograms. It is important to use consistent units when using this equation to solve for weight.

4. Can Kepler's 3rd Law be used for all types of satellites?

Yes, Kepler's 3rd Law can be used for any type of satellite that is orbiting an object. This includes artificial satellites orbiting Earth, natural satellites of other planets, and even objects orbiting stars in our solar system or other galaxies.

5. Are there any limitations to using Kepler's 3rd Law to solve for satellite weight?

While Kepler's 3rd Law is a useful tool for estimating the weight of a satellite, it is not entirely accurate. This is because the law assumes that the orbit is a perfect circle, which is rarely the case in reality. Additionally, the law only takes into account the mass of the object being orbited and does not factor in other forces, such as atmospheric drag, which can affect the weight of a satellite.

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