Question

How high above the surface of Earth is a satellite making four revolutions per day?

Answer 1

to me, I will apply the formula \(T^2 = \dfrac{4\pi^2}{GM}r^3\) to find out the high, where r = \(R_{earth} + distance\)
you have period T can be calculate by 4revolutions /day, that means 1 revolution/ 6hours
T = 6hours = 21600seconds
\[r^3 = \frac{T^2GM}{4\pi^2}\\r =\sqrt[3]{\frac{T^2GM}{4\pi^2}}\]
now replace r = \(R_{earth}+ distance\)
\[distance = \sqrt[3]{\frac{T^2GM}{4\pi^2}}-R_{earth}\]
That's what I think

Answer 2

G = 6.67x10^(-11)
M = 5.972x10^24
R = 6.371x10^6

Answer 3

I think if you change the period into s/rad instead of s/rev you can cancel out the 4pi^2 on the denominator of the cube root radical...

Answer 4

Just working on the signs.\[\sqrt[3]{\frac{T^2GM}{4\pi^2}}\]\[\sqrt[3]{\frac{s^2\cdot \dfrac{N\cdot m^2}{kg^2}kg}{4\pi^2}}\]\[\sqrt[3]{\frac{s^2\cdot \dfrac{kg\cdot m\cdot m^2}{s^2\cdot kg^2}kg}{4\pi^2}}\]\[\sqrt[3]{\frac{\cancel{s^2}\cdot \dfrac{\cancel{kg}\cdot m\cdot m^2}{\cancel{s^2}\cdot \cancel{kg^2}}\cancel{kg}}{4\pi^2}}\]

Answer 5

looks good ^^

Answer 6

I think I'm a dumb - I was thinking about getting the period from the angular frequency - this is just straight 1/f. derpface. sorry :P

Answer 7

I got to the formula through\[F=ma=\frac{mv^2}{r}=\frac{GMm}{r^2}\]\[v^2=\frac{GM}{r}\]\[\left(\frac{2\pi r}{T}\right)^2=\frac{GM}{r}\]@AllTehMaffs How does converting to radians help me from here? Don't I just get \(1/2\pi T\)?

Answer 8

Thanks @Loser66 @AllTehMaffs