The (Earth's) free_fall gravitational_acceleration
 at the Earth's surface (radius  r₁=6378135*[m], equatorial sea_level),
 is (ruffly (approximated (by me) to), simply)
 g₁=~π^{^2}*[m/(s^{^2})]-a_c1
 where
 π^{^2}~9.8
 is multiplied by the (proportionality constant of) units [m/(s^{^2})]
 (& then)
 minus
 its (=the_Earth's) centrifugal_acceleration -a_c1=(v_c1^{^2})/r₁,
 with circumferential_speed v_c1=cir/T=2*π*r₁/T
 for a sidereal_day period T=23*[hr]+56*[min]+4*[s],
 T=23*[hr]*60*[min/hr]*60*[s/min]+56*[min]*60*[s/min]+4*[s],
 T=82800*[s]+3360*[s]+4*[s]
 T=86164*[s]
 & barely produces (a meager loss (from π^{^2}~9.8), of e.g.)
 -0.34% at the equator.

Hovering at higher altitudes
 have more centrifugal_acceleration.

For any (other) height h=r₂-r₁
 above r₁ (sea_level, equatorial)
 the (larger) radius r₂=r₁+h
 is useful.

 g₂=(π^{^2})*[m/(s^{^2})]-a_c2, a_c2=4*(π^{^2})*r₂/(T^{^2})
 g₂=(π^{^2}) [m/(s^{^2})]-(4*(π^{^2}))*r₂/(T^{^2})), bring out π^{^2}
 & move the units into the brackets to (only) factor "1"
 g₂=(π^{^2})*(1 [m/(s^{^2})]-(4*r₂/(T^{^2}))) is too complicated
 but assumes 4*r2/(T^2)=1
 for a hovering, e.g. geo_stationary_orbit GSO (weightlessness),
 or
 g₂=(π^{^2})*(1 [m/(s^{^2})]-r₂*((2/T)^{^2})), let r₂=r₁+h & g₂=g_h

 g₂=(π^{^2})*(1*[m/(s^{^2})]-(r₁+h)*((2/T)^{^2}))   <---!

That (g₂=g_{_h}) is the free_fall acceleration
 wrt height h
 when using
 the sidereal_day period
 T=86164*[s]
 (which is a constant for any height h)
 &
 e.g. the Earth's polar radius
 r₁=6,378,137*[m].

Examples:

Equatorially:

If the height h=0
 at the Earth's equatorial radius (surface)
 r₁=6399592*[m]
 then (the term)
 -(6399592*[m]+h)/(1856058724*[s^{^2}])=-0.00344(646144442)
 which is ~-0.34(46)% (equatorial) loss
 from π^{^2}*[m/(s^{^2})].

Try again, for GSO radius r₂.
 g₂=(π^{^2})*[m/(s^{^2})]-(4*(π^{^2})*r₂/(T^{^2})), +(4*(π^{^2})*r₂/(T^{^2}))
 g₂+(4*(π^{^2})*r₂/(T^{^2}))=(π^{^2})*[m/(s^{^2})], let g₂=0 for a GSOrbit
 4*(π^{^2})*r₂/(T^{^2}))=(π^{^2})*[m/(s^{^2})], *(T^{^2})/(4*(π^{^2}))
 r₂=(T^{^2}/4)*[m/(s^{^2})], T=86164*[s],
 r₂=((86164*[s])^{^2})/4)*[m/(s^{^2})], [s^{^2}]/[s^{^2}]=1/1=1
 r₂=(7,424,234,896)*[m])/4,
 r₂=1,856,058,724*[m]
 is the Earth's GSO radius.

Edited November 29, 20232 yr by Capiert

8 hours ago, Capiert said:

The (Earth's) free_fall gravitational_acceleration
 at the Earth's surface (radius  r₁=6378135*[m], equatorial sea_level),
 is (ruffly (approximated (by me) to), simply)
 g₁=~π^{^2}*[m/(s^{^2})]-a_c1
 where
 π^{^2}~9.8
 is multiplied by the (proportionality constant of) units [m/(s^{^2})]

Pi squared is an acceleration?  Are you sure it isn't a force or a velocity?

12 hours ago, Bufofrog said:

Pi squared is an acceleration?  Are you sure it isn't a force or a velocity?

Pi squared is approximately the magnitude of g. The half-period of a 1m pendulum is very close to 1 sec, and T = 2*pi sqrt(L/g)

(This was a very early realization of the second but is lacking in consistency since g is not constant on the earth’s surface)

—-

g varies with height, so trying to use the surface value for a geostationary satellite is problematic. It’s actually 35,786 km in altitude, or 42,164 km from the center of the earth.

https://en.m.wikipedia.org/wiki/Geostationary_orbit