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

(a). The deflection angle is
`2.77*10^(-9)\ rad`

(b). The deflection angle is
`3.95*10^(-4)\ rad`

(c). The deflection angle is
`7.41*10^(-1)\ rad`

Step-by-step explanation:

Given that,

Mass of earth
`M_(e)=6.0*10^(24)\ kg`

Radius of earth
`R_(e)=6.4*10^(6)\ m`

Mass of white dwarf
`M=2.0*10^(30)\ kg`

Radius of white dwarf
`R=1.5*10^(7)\ m`

Mass of Neutron
`M=3.0*10^(30)\ kg`

Radius of neutron
`R=1.2*10^(4)\ m`

We need to calculate the deflection angle for earth

Using formula of angle

`\alpha=(4G M)/(c^2R)`

Where, R = radius

G = gravitational constant

M = mass

c = speed of light

Put the value into the formula

`\alpha=(4*6.67*10^(-11)*6.0*10^(24))/((3*10^(8))^2*6.4*10^(6))`

`\alpha=2.77*10^(-9)\ rad`

The deflection angle is
`2.77*10^(-9)\ rad`

We need to calculate the deflection angle for white dwarf

Using formula of angle

`\alpha=(4G M)/(c^2R)`

Put the value into the formula

`\alpha=(4*6.67*10^(-11)*2.0*10^(30))/((3*10^(8))^2*1.5*10^(7))`

`\alpha=3.95*10^(-4)\ rad`

The deflection angle is
`3.95*10^(-4)\ rad`

We need to calculate the deflection angle for neutron star

Using formula of angle

`\alpha=(4G M)/(c^2R)`

Put the value into the formula

`\alpha=(4*6.67*10^(-11)*3.0*10^(30))/((3*10^(8))^2*1.2*10^(4))`

`\alpha=7.41*10^(-1)\ rad`

The deflection angle is
`7.41*10^(-1)\ rad`

Hence, This is the required solution.