Fyneman exercise II ch1-1

Q.

Electric and gravitational forces.

a)

What would a proton’s mass be if the gravitational force between two protons at rest were to equal the electric force? How does this compare with its actual mass?

b)

What would be the electric force between two dimes placed at opposite ends of a 10 meter lecture table if their nuclear and electronic charges were unbalanced by about 1%? Can you think of some object whose “weight” equals that force?

A.

symbol	meaning	value
$e$ or $q$	Elementary charge	$1.60 \times 10^{-19}$coulombs
$\epsilon_0$	Vacuum permittivity	$8.85 \times 10^{-12}$ F/m
G	Gravitational constant	$6.67\times10^{-11} m^3\cdot kg^{-1}\cdot s^{-2}$
$m_p$	Proton mass	$1.67 \times 10^{-27}$ kg

a)

Let proton’s mass $M_p$, gravitational force $F_g$, electric force $F_e$ and distance between two protons $r$.

$F_g = G\frac{M_p^2}{r^2}$

$F_e = \frac{1}{4\pi\epsilon_0}\frac{q^2}{r^2}$

Gravitational force equals electric force,

$G\frac{M_p^2}{r^2}=\frac{1}{4\pi\epsilon_0}\frac{q^2}{r^2}$
$\therefore M_p = q\sqrt{\frac{1}{4\pi\epsilon_0G}}$

Using above numbers in table, we get

$M_p = 2 \times 10^{-9}$ kg

$\frac{calculated \ mass} {actual \ mass} = \frac{2 \times 10^{-9}}{1.67\times 10^{-27}} \approx 1.2 \times 10^{18}$

Electric force is very very strong than gravitational force!

b)

Mass of dime = $2.27 \ g= 2.27 \times 10^{-3} \ kg$.

Suppose dime is composed only copper(actually about 91%).

Copper mass is
$63.54 \ u = 63.54\times 1.66\times10^{-27} \ kg= 105.48 \times 10^{-27} \ kg$.

So, dime has about $\frac{2.27\times 10^{-3}}{105.48\times10^{-27}} \approx 0.02 \times 10^{24}$ copper atoms.

$1\%$ of it is $2 \times 10^{20}$.

Force of two dimes is

$\frac{1}{4\pi\epsilon_0}\frac{(2\times 10^{20})^2}{10^2}=\frac{10^{38}}{3.14\times8.85\times 10^{-12}}\approx 3.60\times 10^{52}\ N$

Gravity is $g=9.8 \approx 10 \ m/s^2$.

Object of Mass which equals that force is

$3.60\times 10^{51} \ kg$

May be no such object is on Earth!

Solar Mass is only(?) $5.9722\pm0.0006\times 10^{24}$ kg.