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2 years ago

5

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 ben

t 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

1 answer:

grandymaker [24]2 years ago

3 0

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.

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Derive an expression for the gravitational potential energy of a system consisting of Earth and a brick of mass m placed at Eart

Arlecino [84]

Answer:

The gravitational potential energy of a system is -3/2 (GmE)(m)/RE

Explanation:

Given

mE = Mass of Earth

RE = Radius of Earth

G = Gravitational Constant

Let p = The mass density of the earth is

p = M/(4/3πRE³)

p = 3M/4πRE³

Taking for instance,a very thin spherical shell in the earth;

Let r = radius

dr = thickness

Its volume is given by;

dV = 4πr²dr

Since mass = density* volume;

It's mass would be

dm = p * 4πr²dr

The gravitational potential at the center due would equal;

dV = -Gdm/r

Substitute (p * 4πr²dr) for dm

dV = -G(p * 4πr²dr)/r

dV = -G(p * 4πrdr)

The gravitational potential at the center of the earth would equal;

V = ∫dV

V = ∫ -G(p * 4πrdr) {RE,0}

V = -4πGp∫rdr {RE,0}

V = -4πGp (r²/2) {RE,0}

V = -4πGp{RE²/2)

V = -4Gπ * 3M/4πRE³ * RE²/2

V = -3/2 GmE/RE

The gravitational potential energy of the system of the earth and the brick at the center equals

U = Vm

U = -3/2 GmE/RE * m

U = -3/2 (GmE)(m)/RE

5 0

1 year ago

a block of mass m slides along a frictionless track with speed vm. It collides with a stationary block of mass M. Find an expres

shusha [124]

Answer:

Part a) When collision is perfectly inelastic

$v_m = \frac{m + M}{m} \sqrt{5Rg}$

Part b) When collision is perfectly elastic

$v_m = \frac{m + M}{2m}\sqrt{5Rg}$

Explanation:

Part a)

As we know that collision is perfectly inelastic

so here we will have

$mv_m = (m + M)v$

so we have

$v = \frac{mv_m}{m + M}$

now we know that in order to complete the circle we will have

$v = \sqrt{5Rg}$

$\frac{mv_m}{m + M} = \sqrt{5Rg}$

now we have

$v_m = \frac{m + M}{m} \sqrt{5Rg}$

Part b)

Now we know that collision is perfectly elastic

so we will have

$v = \frac{2mv_m}{m + M}$

now we have

$\sqrt{5Rg} = \frac{2mv_m}{m + M}$

$v_m = \frac{m + M}{2m}\sqrt{5Rg}$

6 0

2 years ago

A car travels 10 m/s east. Another car travels 10 m/s north. The relative speed of the first car with respect to the second is:

Thepotemich [5.8K]

Answer:

d. less than 20m/s

Explanation:

To the 2nd car, the first car is travelling 10m/s east and 10m/s south. So the total velocity of the first car with respect to the 2nd car is

[tex]\sqrt{10^2 + 10^2} =10\sqrt{2}=14.14m/s

As 14.14m/s is less than 20m/s. d is the correct selection for this question.

3 0

1 year ago

Suppose you are talking by interplanetary telephone to your friend, who lives on the Moon. He tells you that he has just won a n

Savatey [412]

Answer:

The friend on moon will be richer.

Explanation:

We must calculate the mass of gold won by each person, to tell who is richer. For that purpose we will use the following formula:

W = mg

m = W/g

where,

m = mass of gold

W = weight of gold

g = acceleration due to gravity on that planet

<u>FOR FRIEND ON MOON</u>:

W = 1 N

g = 1.625 m/s²

Therefore,

m = (1 N)/(1.625 m/s²)

m(moon) = 0.6 kg

<u>FOR ME ON EARTH</u>:

W = 1 N

g = 9.8 m/s²

Therefore,

m = (1 N)/(9.8 m/s²)

m(earth) = 0.1 kg

Since, the mass of gold on moon is greater than the mass of moon on earth.

<u>Therefore, the friend on moon will be richer.</u>

7 0

2 years ago

The mass m1 enters from the left with velocity v0 and strikes a mass m2 > m1 which is initially at rest. The collision betwee

enot [183]

Answer:

1. False 2) greater than. 3) less than 4) less than

Explanation:

1)

As the collision is perfectly elastic, kinetic energy must be conserved.
The expression for the final velocity of the mass m₁, for a perfectly elastic collision, is as follows:

$v_{1f} = v_{10} *\frac{m_{1} -m_{2} }{m_{1} +m_{2}}$

As it can be seen, as m₁ ≠ m₂, v₁f ≠ 0.

2)

As total momentum must be conserved, we can see that as m₂ > m₁, from the equation above the final momentum of m₁ has an opposite sign to the initial one, so the momentum of m₂ must be greater than the initial momentum of m₁, to keep both sides of the equation balanced.

3)

The maximum energy stored in the in the spring is given by the following expression:

$U =\frac{1}{2} *k * A^{2}$

where A = maximum compression of the spring.

This energy is always the sum of the elastic potential energy and the kinetic energy of the mass (in absence of friction).
When the spring is in a relaxed state, the speed of the mass is maximum, so, its kinetic energy is maximum too.
Just prior to compress the spring, this kinetic energy is the kinetic energy of m₂, immediately after the collision.
As total kinetic energy must be conserved, the following condition must be met:

$KE_{10} = KE_{1f} + KE_{2f}$

So, it is clear that KE₂f < KE₁₀
Therefore, the maximum energy stored in the spring is less than the initial energy in m₁.

4)

As explained above, if total kinetic energy must be conserved:

$KE_{10} = KE_{1f} + KE_{2f}$

So as kinetic energy is always positive, KEf₂ < KE₁₀.

4 0

1 year ago