Asked by Ana

A satellite of mass 220 kg is launched from a site on Earth's equator into an orbit at 200 km above the surface of Earth.
(a) Assuming a circular orbit, what is the orbital period of this satellite?
5310 s

(b) What is the satellite's speed in its orbit?
7790m/s

(c) What is the minimum energy necessary to place the satellite in orbit, assuming no air friction?

I was able to get a and b, but i don't know about c.

Thanks
Answered by Damon
You have to lift the body to orbit and then speed it up to orbit speed.
m v^2/r = G m M/r^2 in orbit
so (1/2)m v^2 = G m M/(2r) = Ke in orbit
Change in Potential energy to lift it to orbit
= G m M/(r-Rearth)
Answered by Ana
how do i get small r?
Answered by Ana
is it R_e + height?
Answered by Damon
Yes, earth radius + height
Answered by Ana
(6.67E-11*220*5.98E24)/(6.38E6+2.0E5)-(6.38E6)=
1.33295653E10J

Is this right? I want to make sure if this is right before I insert into the site.
Answered by Damon
both terms should be positive
Answered by Ana
What do you mean?
Answered by Damon
(6.67E-11*220*5.98E24)/[2(6.38E6+2.0E5)]
note 2 in denominator
+ (6.67E-11*220*5.98E24) / 2^10^5
Answered by Damon
8.78*10^16 [ 1/1.32E7 + 1/2E5 ]
Answered by Ana
so is it 6770873457?
Answered by Damon
Change in Potential energy to lift it to orbit
= G m M [1/Rearth - 1/r ]
Answered by Damon
8.78*10^16 [ 1/1.32E7 + 1/6.38E6 - 1/6.58E6 ]
Answered by Ana
this is my last try....is it 6904053501?
Answered by Damon
7.07*10^9 Joules
Answered by Damon
We are close to each other. My arithmetic is failing, after bedtime.
Answered by Damon
Your professor must like arithmetic !
Answered by Ana
Yes, we are. Some of the problems are very difficult but thank you so much for your help. I really appreciate.
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