What is the Height of the International Space Station Above the Earth's Surface?

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The discussion revolves around calculating the height of the International Space Station (ISS) above Earth's surface, which orbits at 15.65 revolutions per day. The initial calculations used standard gravitational acceleration (9.8 m/s²), but this approach is flawed due to the weakening of gravity at higher altitudes. Participants suggest using the formula for gravitational force and consider Kepler's third law for a more accurate solution. The correct height of the ISS is approximately 400 kilometers, significantly lower than the initial erroneous calculation of 1196 kilometers. The conversation emphasizes the importance of adjusting gravitational calculations for altitude.
Chan M

Homework Statement


The International Space Station, launched in 1998, makes 15.65 revolutions around the Earth each day in a circular orbit. Find the height of the space station (in kilometers) above the earth’s surface.

Homework Equations


T = (2*pi*r) / v
r - radius, distance between center of Earth and the station
r = Re + h
Re - radius of earth
h = height above ground
Find h

The Attempt at a Solution


So there is only one force acting on the station, which is gravity. Acceleration due to gravity is also centripetal acceleration. Gravity is the centripetal force.

15.65 revolution each day is one revolution every 5520.77 seconds

Fg = m*Ac
mg = m * (v^2 / r)
g = v^2 / r
g = [ (2*pi*r) / T ]^2 / r
g = 4*pi*pi*r / T^2
h = [(T^2)*g / 4*pi*pi ] - Re
h = [(5520.77^2)*9.8 / 4*pi*pi ] - (6.37 * 10^6)
h =
upload_2017-10-2_21-38-59.png
(this is what I put in the calculator)
h = 1195987.99 meters
h = 1196 km

This is not the correct answer, I know the correct answer, I just don't know how to get it or what I'm doing wrong. Thanks!
 

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Gravity is weaker that high up so ##g=9.8##m/s won't work. You need
$$g=G\frac{M_{Earth}}{r^{2}}$$
 
can one use Kepler's 3rd law?
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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