# What is the K.E. of the satellite?

• starchaser2020
In summary, a 5,000.0 kg satellite has been placed in a circular, sun-synchronous orbit at an altitude of 1300.0 km. The K.E. of the satellite can be calculated using the formula K=(1/2)mv^2. The forces acting on the satellite in that orbit and its acceleration may not be affected by the sun synchronicity.

#### starchaser2020

A satellite has been placed in a circular, sun-synchronous orbit at an altitude of 1300.0 km. The satellite has a mass of 5,000.0 kg. What is the K.E. of the satellite?

I know K=(1/2)mv^2 ... but I have no idea where to go from here. I Know I am over-thinking this. So any help would be greatly appreciated!

What forces act on the satellite in that orbit? What is its acceleration?
(I don't think the sun-synchronicity needs to be taken into account.)

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PeroK said:
The sun synchronicity is telling you how long the satellite takes to orbit. You could actually work this out using the radius if you know enough about circular motion, as suggested by haruspex above.

But I suspect you won't know enough physics yet so you have been given the period of orbit.

How long does the sun take to "orbit" the earth?
Interesting... it's not a term I'd come across, so I looked it up before answering the post. My reading is that sun synchronicity is not directly related to orbital period. Indeed, I read that typical sun synchronous orbits are at around 500-600km and take under two hours. The term refers to the orbits being so arranged that the Earth's oblateness causes the orbit to precess one day per year, thereby always appearing the same from the sun's perspective. Hence it didn't seem relevant to me.

Yes, you're right. It's not about the period at all. Apologies.