From Newsgroup: sci.math
The 'nym-shifting troll trolled as "Efrawin Muksunov":
Paul B. Andersen wrote:
Den 24.01.2026 07:51, skrev Maciej Wo+|niak:
The vertical component of the force acting on the object is:
-a-a Fb|N = 9.8 N-a (constant upwards)
So what?
Do you have a reading comprehension problem?
So according to Newton's law F = ma the acceleration of the object is a
= F/m = 9.8 m/s-# (upwards)
idiot, that's g = F/m = 9.8 m/s-#, not a.
JFTR:
g is just a variable that is used to emphasize that the acceleration is a gravitational acceleration.
In this aspect, a constant , not acceleration
Wrong. First of all, that an acceleration is constant does not mean that
the associated velocity is constant, too. If the acceleration is constant, A(t) = A(0), but not zero, then the associated velocity is NOT constant:
A(t) = dV(t)/dt ==> V(t) = int dt A(t) = int dt A(0) = V(0) + A(0) t.
Also, the gravitational acceleration (field) is not actually constant in
*any* way. That is merely an approximation that only works for short
periods of time and small distances in special cases; in particular, it
works for assumed standalone spherical celestial bodies with uniform mass distribution, for locations outside the body, and altitudes (measured from their surface) that are small compared to their radii.
In general, for such bodies one has from Newton's Second Law of Motion, and Newton's Law of Universal Gravitation:
F(R) = m_i A(R) = -(G M m_g/r^2) R/r = m_g g(R) =: F_g(R),
where R is a radius vector, r is its magnitude (the distance from the center
of mass), m_i is inertial mass and m_g is gravitational mass.
One can then identify the gravitational acceleration to be
A(R) = g(R) = -(m_g/m_i) (G M/r^2) R/r.
The negative sign indicates that it points inwards (towards the origin where one defines the geometrical center of the celestial object to be) as the
radius vector points outwards.
With the assumption m_i = m_g (weak equivalence principle, experimentally confirmed to high precision), one has
g(R) = -(G M/r^2) R/r.
One can thus see that the gravitational acceleration varies with position:
at equal radii, its direction varies; and at equal angular coordinates, its magnitude varies with the radius:
g(r) := ||g(R)||_2 = G M/r^2.
[Notice that g(r) is NOT singular at r = 0 because one must substitute the effective mass M := M(r < R) = rho V(r < R) = (4pi/3) r^3 rho, where R is
now the radius of the body, and rho = M/V is the assumed constant mass density.]
The distance from the center of mass (which due to the spherical symmetry
and uniform mass density is identical to the geometrical center of the celestial object) is the sum of the radius and the altitude:
r = R + h,
so
g(h) = G M/(R + h)^2
= G M/(R^2 + 2 R h + h^2)
= G M/[R^2 (1 + 2 h/R + h^2/R^2).
When the (difference in) altitude is small compared to the radius, i.e.
h/R =~ 0, then the gravitational acceleration is approximately constant
(as long as the mass of the celestial body and its radius are constant):
g =~ G M/R^2.
For Terra, we have M =~ 5.97 * 10^24 kg (estimated e.g. from orbits of
natural and artificial satellites; see also: Cavendish experiment) and
R =~ 6371 km (average radius), which leads to the average magnitude of
the terrestrial surface gravitational acceleration
g =~ 9.82 m/s^2.
Therefore the standard magnitude of the terrestrial surface gravitational acceleration is *defined* to be
g_0 = 9.8 m/s^2.
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