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Helga [31]
3 years ago
9

When does gravitational lensing occur? High concentrations of dark matter cause length contraction of nearby objects. The gravit

y from a distant object (like a quasar) distorts the appearance of nearby galaxies. Telescope lenses are distorted by gravitational forces from dark matter. A massive object (like a galaxy cluster) bends the light from an object (like a quasar) that lies behind it.
Physics
1 answer:
Stella [2.4K]3 years ago
4 0

Answer:

A massive object (like a galaxy cluster) bends the light from an object (like a quasar) that lies behind it.

Explanation:

A massive object, like a galaxy cluster, is able to deform the space-time shape as a consequence of its own gravity, so the light that it is coming from a source that is behind it in the line of sight will be bend or distorts in a way that will be magnified, making small arcs around the cluster with the image of the background object.

This technique is useful for astronomers since they make research of faraway objects (at hight redshift) that otherwise will difficult to detect with a telescope.

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A satellite in a circular orbit of radius R around planet X has an orbital period T. If Planet X had one-fourth as much mass, th
Iteru [2.4K]
<h2>Answer: 2T</h2>

According to the Third Kepler’s Law of Planetary motion <em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.</em>

In other words, this law states a relation between the orbital period T of a body (moon, planet, satellite) orbiting a greater body in space with the size R of its orbit.

This Law is originally expressed as follows (in the case of planet X and assuming we have a circular orbit):

T^{2}=\frac{4\pi^{2}}{GM}R^{3}    (1)

Where:

G is the Gravitational Constant

M=1.9(10)^{27}kg is the mass of planet X

R  is the radius of the orbit of the satellite around planet X

If we want to find the period, we have to express equation (1) as written below and substitute all the values:

T=2\pi\sqrt{\frac{R^{3}}{GM}}   (2)

Now, we are asked to find the period when tha mass of the planet is \frac{1}{4}M. In order to do this, we have to rewrite equation (2) with this new value:

T=2\pi\sqrt{\frac{R^{3}}{G(\frac{1}{4}M)}}  (3)

Solving:

T=4\pi\sqrt{\frac{R^{3}}{G(\frac{1}{4}M)}}   (4)

On the other hand, if we multiply both sides of equation (2) by 2, we have:

2T=4\pi\sqrt{\frac{R^{3}}{GM}}    (5)

As we can see, (5) is equal to (4). This means the orbital period is twice the orignal period.

Hence, the answer is:

If Planet X had <u>one-fourth </u>as much mass, the <u>orbital period</u> of this satellite in an orbit of the same radius would be <u>2T.</u>

3 0
3 years ago
A 20 g bullet is shot from a 10 kg gun with a velocity of 400 m/s. What
Nady [450]

Answer: -.80m/s

Explanation:

3 0
3 years ago
Which two types of air masses would likely form a subtropical jet stream?
shepuryov [24]
The exact location of the 'subtropical-jet stream' is located at the North phase of 30 Degrees and the reason behind this is due to the variation of air which lies on the region of mid-altitude and warmer equatorial air. the correct answer would be Cool and Warm air masses meeting near the equator.
5 0
2 years ago
Read 2 more answers
After a package is dropped from the plane, how long will it take for it to reach sea level from the time it is dropped? assume t
Ivenika [448]

Time taken by the package to reach the sea level= 13.7 s

height=h=925 m

initial velocity along vertical= vi=0

acceleration due to gravity=g=9.8 m/s^2

using the kinematic equation h= Vi*t + 1/2 gt^2

925=0(t)+1/2 (9.8)t^2

4.9 t^2=925

t= 13.7 s

6 0
3 years ago
A body in simple harmonic motion has a displacement x that varies in time t according to the equation x = 5cos(π t + π/3) , wher
zhuklara [117]

Answer:

1/2 Hz

Explanation:

A simple harmonic motion has an equation in the form of

x(t) = Acos(\omega t - \phi)

where A is the amplitude, \omega = 2\pi f is the angular frequency and \phi is the initial phase.

Since our body has an equation of  x = 5cos(π t + π/3) we can equate \omega = \pi and solve for frequency f

2\pi f = \pi

f = 1/2 Hz

7 0
3 years ago
Read 2 more answers
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