Question

A light ray just grazes the surface of the Earth (M = 6.0 × 10 24 kg, R = 6.4×10 6 m). Through what angle α is the light ray bent by gravitational lensing? (Ignore the refractive effects of the Earth’s atmosphere.) Repeat your calculation for a white dwarf (M = 2.0 × 10 30 kg, R = 1.5 × 10 7 m) and for a neutron star (M = 3

Answer 1

Answer:

(a). The deflection angle is $2.77\times10^{-9}\ rad$

(b). The deflection angle is  $3.95\times10^{-4}\ rad$

(c). The deflection angle is $7.41\times10^{-1}\ rad$

Explanation:

Given that,

Mass of earth $M_{e}=6.0\times10^{24}\ kg$

Radius of earth $R_{e}=6.4\times10^{6}\ m$

Mass of white dwarf $M=2.0\times10^{30}\ kg$

Radius of white dwarf $R=1.5\times10^{7}\ m$

Mass of Neutron $M=3.0\times10^{30}\ kg$

Radius of neutron $R=1.2\times10^{4}\ m$

We need to calculate the deflection angle for earth

Using formula of angle

$\alpha=\dfrac{4G M}{c^2R}$

Where, R = radius

G = gravitational constant

M = mass

c = speed of light

Put the value into the formula

$\alpha=\dfrac{4\times6.67\times10^{-11}\times6.0\times10^{24}}{(3\times10^{8})^2\times6.4\times10^{6}}$

$\alpha=2.77\times10^{-9}\ rad$

The deflection angle is $2.77\times10^{-9}\ rad$

We need to calculate the deflection angle for white dwarf

Using formula of angle

$\alpha=\dfrac{4G M}{c^2R}$

Put the value into the formula

$\alpha=\dfrac{4\times6.67\times10^{-11}\times2.0\times10^{30}}{(3\times10^{8})^2\times1.5\times10^{7}}$

$\alpha=3.95\times10^{-4}\ rad$

The deflection angle is $3.95\times10^{-4}\ rad$

We need to calculate the deflection angle for neutron star

Using formula of angle

$\alpha=\dfrac{4G M}{c^2R}$

Put the value into the formula

$\alpha=\dfrac{4\times6.67\times10^{-11}\times3.0\times10^{30}}{(3\times10^{8})^2\times1.2\times10^{4}}$

$\alpha=7.41\times10^{-1}\ rad$

The deflection angle is $7.41\times10^{-1}\ rad$

Hence, This is the required solution.