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On the other hand, a balance, such as a typical double-pan balance, compares the gravitational force (weight) on the unknown mass to the gravitational force on some reference mass. …

This label is called inertial mass. … of the reference particle, $$m=\frac{|\vec a_0|}{|\vec a|}m_0.$$ The other mass to be defined in classical mechanics is gravitational mass. …

This is a quite subtle problem. You have to be careful about three different situations. A ball can be thrown with velocity (relative to the ground): a) $v_0-v_e$. b) $v(t)-v_e$, where $v(t)$ is the …

The gravitational mass, $m_g$, gives you the strength of the gravitational interaction while the inertial mass, $m_i$, represents the inertia of the body. … The first one is the mass appearing in the Universal Gravitation Law while the second one is the mass appearing in the Newton's second law. …

However, the difference must be small due to the small mass of the Moon. …

$M$ stands for the mass of the Sun. $h$ is the angular momentum. It is defined as $\vec h=\vec r\times\vec p$, where $\vec p$ is the linear momentum. … They are related as $$A=\frac{hT}{2m},$$ where $m$ is the mass of the planet. $r_{min}$ and $r_{max}$ are minimum and maximum distance from the planet to the Sun. …

The $M+m$ in third Kepler's law is a vestige of the reduced mass associated to the two body problem. … speaking we map a coupled and complicated system of two interacting particles into an equivalent problem of decoupled differential equations, one of them describing the motion of a particle of reduced mass …

means the mass of a particle is an intrinsic property. … In any of these cases you will get equations of motions with gravitational mass on one side and inertial mass on the other. In principle there would be no reason for them to be equal. …

If one considers a particle of mass $m$ and initial velocity $v_1$ striking a target of mass $m'$ at rest, without changing its direction, then its final velocity $v_2$ can assume two possible values, … By the time, the mass of the electron was known to be much smaller than the mass of the alpha particle so a backscattering event would imply that the scattering centers were in fact heavy positive nuclei …

For any finite thickness we can consider a layer of mass whose superficial density is $\sigma$. … The gravitational constant is $G=6.67\times 10^{-11}\, ~\mathrm{Nm^2/kg}.$ If the layer of mass is $d$ meters thick and made of a material with the same mean mass density of the Earth ($\rho=5.5\cdot 10 …

The eigenvalues of $P_\mu P^\mu$ are $m^2$, the square mass of the particle. This gives rise to an infinite dimensional representation whose states are labeled by four momentum $p$. …