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3 years ago

What would happen to a satellite if it’s orbit speed was reduced by half? Also how about if it doubled?

1 answer:

7 0

For finding the orbital speed of the satellite we can say that the centripetal force for the circular motion of satellite is provided by the gravitational force of earth

so here we can say

$F_g = \frac{mv^2}{r}$

$\frac{GMm}{r^2} = \frac{mv^2}{r}$

now we will have

$v = \sqrt{\frac{GM}{r}}$

now here we will say that orbital speed of the satellite is inversely depends on the orbital radius

So here if orbital speed is half then as per above relation we can say that orbital distance will become four times

Also we can say that if orbital speed is double then orbital distance will become one fourth of initial distance.

You might be interested in

At the instant shown in the diagram, the car's centripetal acceleration is directed

serg [7]

Answer:

At the instant shown in the diagram, the car's centripetal acceleration is directed is discussed below in detail.

Explanation:

The direction of the centripetal acceleration is in a circular movement is forever towards the middle of the roundabout pathway. In the picture displayed, the East direction is approaching the center. So, the course of the car's centripetal acceleration is (H) toward the east.

3 0

3 years ago

An extremely long wire laying parallel to the x -axis and passing through the origin carries a current of 250A running in the po

viktelen [127]

Answer:

$1.232\times 10^{-5}\ T.$

Explanation:

Given:

Current through the wire, passing through the origin, $I_1 = 250\ A.$

Current through the wire, passing through the y axis, $r_y=1.8\ m.$ , $I_2 = 50\ A.$

According to Ampere's circuital law, the line integral of magnetic field over a closed loop, called Amperian loop, is equal to $\mu_o$ times the net current threading the loop.

$\oint \vec B \cdot d\vec l=\mu_o I.$

In case of a circular loop, the directions of magnetic field and the line element $d\vec l$ , both are along the tangent of the loop at that point, therefore, $\vec B\cdot d\vec l = B\ dl$ .

$\oint \vec B \cdot d\vec l = \oint B\ dl = B\oint dl.\\$

$\oint dl$ is the circumference of the Amperian loop = $2\pi r$

Therefore,

$B\ 2\pi r=\mu_o I\\B=\dfrac{\mu_o I}{2\pi r}.$

It is the magnetic field due to a current carrying wire at a distance r from it.

For the first wire, passing through the origin:

Consider an Amperian loop of radius 3.510 m, concentric with the axis of the wire, such that it passes through the point where magnetic field is to be found, therefore, $r_1 = 3.510\ m.$

The magnetic field at the given point due to this wire is given by:

$B_1 = \dfrac{\mu_o I_1}{2\pi r_1}\\=\dfrac{4\pi \times 10^{-7}\times 250}{2\pi \times 3.510}=1.42\times 10^{-5}\ T.$

For the first wire, passing through the y-axis:

Consider an Amperian loop of radius (3.510+1.8) m = 5.310 m, concentric with the axis of the wire, such that it passes through the point where magnetic field is to be found, therefore, $r_2 = 5.310\ m.$

The magnetic field at the given point due to this wire is given by:

$B_2 = \dfrac{\mu_o I_2}{2\pi r_2}\\=\dfrac{4\pi \times 10^{-7}\times 50}{2\pi \times 5.310}=1.88\times 10^{-6}\ T.$

The directions of current in both the wires are opposite therefore, the directions of the magnetic field due to both the wires are also opposite to that of each other.

Thus, the net magnetic field at $r_r=-3.510\ m$ is given by

$B=B_1-B_2 = 1.42\times 10^{-5}-1.88\times 10^{-6}=1.232\times 10^{-5}\ T.$

6 0

3 years ago

During a solar eclipse, the Moon, Earth, and Sun all lie on the same line, with the Moon between the Earth and the Sun.

lord [1]

Answer:

(a) $F_{sm} = 4.327\times 10^{20}\ N$

(b) $F_{em} = 1.983\times 10^{20}\ N$

Solution:

As per the question:

Mass of Earth, $M_{e} = 5.972\times 10^{24}\ kg$

Mass of Moon, $M_{m} = 7.34\times 10^{22}\ kg$

Mass of Sun, $M_{s} = 1.989\times 10^{30}\ kg$

Distance between the earth and the moon, $R_{em} = 3.84\times 10^{8}\ m$

Distance between the earth and the sun, $R_{es} = 1.5\times 10^{11}\ m$

Distance between the sun and the moon, $R_{sm} = 1.5\times 10^{11}\ m$

Now,

We know that the gravitational force between two bodies of mass m and m' separated by a distance 'r' is given y:

$F_{G} = \frac{Gmm'_{2}}{r^{2}}$ (1)

Now,

(a) The force exerted by the Sun on the Moon is given by eqn (1):

$F_{sm} = \frac{GM_{s}M_{m}}{R_{sm}^{2}}$

$F_{sm} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 7.34\times 10^{22}}{(1.5\times 10^{11})^{2}}$

$F_{sm} = 4.327\times 10^{20}\ N$

(b) The force exerted by the Earth on the Moon is given by eqn (1):

$F_{em} = \frac{GM_{s}M_{m}}{R_{em}^{2}}$

$F_{em} = \frac{6.67\times 10^{- 11}\times 5.972\times 10^{24}\times 7.34\times 10^{22}}{(3.84\times 10^{8})^{2}}$

$F_{em} = 1.983\times 10^{20}\ N$

$F_{se} = \frac{GM_{s}M_{m}}{R_{es}^{2}}$

$F_{se} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 5.972\times 10^{24}}{((1.5\times 10^{11}))^{2}}$

$F_{se} = 3.521\times 10^{20}\ N$

7 0

3 years ago

Calculate the resistance of the following network by R=R1+R2+R3

gayaneshka [121]

Answer:

R=60Ω

Explanation:

You are looking for the total resistance of a parallel circuit so the formula is

R=R1+R2+R3, where R1=10Ω, R2=20Ωand R3=30Ω

R=10Ω+20Ω+30Ω

R=60Ω

5 0

3 years ago

A billiard ball of mass m hits another one of the same mass. The first ball moves off at 30 degrees. For an elastic collision wh

sleet_krkn [62]

Answer:

v₁ = u₁/2√3 ≈ 0.866u₁

v₂ = u₁/2 = 0.5u₁

θ = 60°

Explanation:

let u₁ be the initial velocity of the first ball

let v₁ be the final velocity of the first ball

let v₂ be the final velocity of the second ball

For elastic collisions, the angle between the departing masses is 90°

assume the first ball initially moves along the x axis in the positive direction

conservation of momentum

In the y direction, initial momentum is zero

After the collision

mv₁sin30 = mv₂sin60

½v₁ = ½√(3)v₂

v₁ = √(3)v₂

in the x direction,

mu₁ = mv₁cos30 + mv₂cos-60

u₁ = v₁cos30 + v₂cos60

u₁ = (√(3)v₂)½√(3) + ½v₂

u₁ = 2v₂

v₂ = u₁/2

v₁ = √(3)v₂ = √(3)(u₁/2)

6 0

3 years ago