# Question #8eda3

Apr 5, 2016

See two situations below.

#### Explanation:

• Your conjecture that a particle will fall from infinity towards earth is not supported by the Newton's Law of Gravitation which gives that force of attraction and between earth having mass ${M}_{e}$, distance between the centers of earth and the object $R$ and mass of object being $m$ is given by the expression

${F}_{G} = G \frac{{M}_{e} . m}{R} ^ 2$

Where $G$ is the proportionality constant and has the value $6.67408 \times {10}^{-} 11 {m}^{3} k {g}^{-} 1 {s}^{-} 2$

Since the particle is situated at $\infty$,
${F}_{G} = G \frac{{M}_{e} . m}{R} ^ 2 = G \frac{{M}_{e} . m}{\infty} ^ 2 = 0$.
There is no force of attraction and as such there is no reason that particle should fall towards earth.

• Hypothetically if we assume that the particle is having velocity $v$ and it somehow moves toward earth. Without this assumption there is no movement of the particle at all.
Its gravitational potential energy with respect to earth at infinity$= - G {M}_{e} \frac{m}{R} = 0$
On reaching the surface of earth let its velocity be ${v}_{f}$. Assuming non-relativistic velocities, and applying Law of Conservation of Energy

$\text{Initial Gravitational PE"+"Initial KE"="Final Gravitational PE"+"Final KE}$
$\implies 0 + \frac{1}{2} m {v}^{2} = - G {M}_{e} \frac{m}{R} _ e + \frac{1}{2} m {v}_{f}^{2}$,
where ${R}_{e}$ is the radius of earth.

$\implies \frac{1}{2} m {v}_{f}^{2} = \frac{1}{2} m {v}^{2} + G {M}_{e} \frac{m}{R} _ e$
$\implies {v}_{f} > v$
This is a fallacy. We have created energy by moving the particle from infinity to surface of earth without doing any work.

It implies that the hypothesis is not supported by the know laws of Physics, as such it is incorrect.
To me it appears that only first situation is correct.