At what altitude above equator do gravitational and centrifugal forces cancel each other? [closed]

Question

How high has to be a tower at equator to get on its peak the same - but of course inverted sing - value of gravitational and centrifugal forces?

Answer 1

If you are asking the question "How high does a person have to be in order to be "weightless" because gravity is canceled from the rotation of the Earth?" as paraphrased by CoilKid, then what you are talking about is a Geostationary Orbit. You can only achieve this orbit in a equatorial plane for the orbitting object to seem truly stationary with respect to an observer on Earth, because all circular orbits are centred on the Earth's centre. The orbit's radius is about 36 000km above the Earth's equator, i.e. about 42 000km from the centre of the Earth. You can't keep something stationary with respect to a non-equatorial observer: if you were looking at the object from Regensburg, it would come back to Regensburg at the same time each siderial day (look up the definition of sidereal day - it would seem to come back a little earlier each day), but its latitude would oscillate between 49 degrees North and South each sidereal day. You would see it oscillating in roughly a North-South plane (actually it is a figure of eight), roughly sinusoidally with time. This is a geosynchronous orbit, rather than geostationary. You calculate the altitude height $r$ above centre of Earth from the assertion that the local acceleration owing to gravity equals the centrepetal force of an orbit with a period of one sidereal dat $T_s$:

$$\frac{G\,M_\oplus}{r^2} = \omega_s^2 r$$

where $\omega_s = 2\pi/T_s$ is $2\pi$ radians per sidereal day $T_s\approx 23.9\times 3\,600{\rm s}$ and $M_\oplus$ the Earth's mass. You can simplify this in terms of known $g$, since $G\,M/r_\oplus^2 = g$, as:

$$g\,\frac{r_\oplus^2}{r^2} = \omega_s^2 r$$

whence I reckon $r$ to be (taking $r_\oplus\approx 6\,400{\rm km}$:

$$r=\sqrt[3]{g\,\frac{r_\oplus^2\,T_s^2}{4\,\pi^2}} = 42\, 270{\rm km}$$

i.e. an altitude of $ 42\, 270-6\,400 \approx 36\,000{\rm km}$.

Answer 2

Think about this. If gravity is of equal magnitude as the centrifugal force, a satellite there will have an angular velocity same as the earth rotating angular velocity. However, this is just a geosynchronous satellite!

The equation is $$\frac{GMm}{r^2}=mr\frac{4\pi^2}{T^2}$$ where $T=24h$.

Hope it helps!