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Alexxx [7]
3 years ago
14

Planet 1 orbits Star 1 and Planet 2 orbits Star 2 in circular orbits of the same radius. However, the orbital period of Planet 1

is longer than the orbital period of Planet 2. What could explain this
Physics
1 answer:
hichkok12 [17]3 years ago
6 0

Answer:

The mass of Star 2 is Greater than the mass of Start 1. (This, if we suppose the masses of the planets are much smaller than the masses of the stars)

Explanation:

First of all, let's draw a free body diagram of a planet orbiting a star. (See attached picture).

From the free body diagram we can build an equation with the sum of forces between the start and the planet.

\sum F=ma

We know that the force between two bodies due to gravity is given by the following equation:

F_{g} = G\frac{m_{1}m_{2}}{r^{2}}

in this case we will call:

M= mass of the star

m= mass of the planet

r = distance between the star and the planet

G= constant of gravitation.

so:

F_{g} =G\frac{Mm}{r^{2}}

Also, if the planet describes a circular orbit, the centripetal force is given by the following equation:

F_{c}=ma_{c}

where the centripetal acceleration is given by:

a_{c}=\omega ^{2}r

where

\omega = \frac{2\pi}{T}

Where T is the period, and \omega is the angular speed of the planet, so:

a_{c} = ( \frac{2\pi}{T})^{2}r

or:

a_{c}=\frac{4\pi^{2}r}{T^{2}}

so:

F_{c}=m(\frac{4\pi^{2}r}{T^{2}})

so now we can do the sum of forces:

\sum F=ma

F_{g}=ma_{c}

G\frac{Mm}{r^{2}}=m(\frac{4\pi^{2}r}{T^{2}})

in this case we can get rid of the mass of the planet, so we get:

G\frac{M}{r^{2}}=(\frac{4\pi^{2}r}{T^{2}})

we can now solve this for T^{2} so we get:

T^{2} = \frac{4\pi ^{2}r^{3}}{GM}

We could take the square root to both sides of the equation but that would not be necessary. Now, the problem tells us that the period of planet 1 is longer than the period of planet 2, so we can build the following inequality:

T_{1}^{2}>T_{2}^{2}

So let's see what's going on there, we'll call:

M_{1}= mass of Star 1

M_{2}= mass of Star 2

So:

\frac{4\pi^{2}r^{3}}{GM_{1}}>\frac{4\pi^{2}r^{3}}{GM_{2}}

we can get rid of all the constants so we end up with:

\frac{1}{M_{1}}>\frac{1}{M_{2}}

and let's flip the inequality, so we get:

M_{2}>M_{1}

This means that for the period of planet 1 to be longer than the period of planet 2, we need the mass of star 2 to be greater than the mass of star 1. This makes sense because the greater the mass of the star is, the greater the force it applies on the planet is. The greater the force, the faster the planet should go so it stays in orbit. The faster the planet moves, the smaller the period is. In this case, planet 2 is moving faster, therefore it's period is shorter.

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As more and more resistors are added in parallel to a circuit, the equivalent resistance of the circuit ____________ (increases,
Anestetic [448]

Answer:

Decreases, Increases

Explanation:

Resistance is parallel can be calculated using

1/Req = 1/R1 + 1/R2 + 1/R3 +....

Then, as more resistor is added in parallel the equivalent resistance is reduced.  

Let use a simple sample

Let all the resistor have equal resistances

Let say R = R1 = R2 = R3 =...Rn

Then, 1/Req = 1/R1 + 1/R2 + 1/R3 +....

1/Req = 1/R + 1/R + 1/R +.... 1/Rn

Req = R/n

Check attachment on how I got that.

This implies that, the equivalent resistance will always be less than the original resistance, since n>1

So, as n increases (I.e. as the number of resistance increases), the equivalent resistance reduces.

B. Now, to know if the current reduces or increases

Using Ohms law

V = iR

Then, I = V/R

So, let assume the voltage is constant, then, the current is inversely proportional to the resistance, so as we know that the resistance is reducing, then the current will be increasing.

So current increase as we add more resistor in parallel to a circuit

3 0
3 years ago
Giving lots of points
bekas [8.4K]

Answer:

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Explanation:

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san4es73 [151]
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Where a subducting plate slides beneath the lithosphere, melting takes place and a(n) ____ is created?
11111nata11111 [884]

under ground cave will be created


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Peg P is driven by the forked link OA along the path described by r = eu, where r is in meters. When u = p4 rad, the link has an
8_murik_8 [283]

Answer:

The transverse component of acceleration is 26.32 m/s^2 where as radial the component of acceleration is 8.77 m/s^2

Explanation:

As per the given data

u=π/4 rad

ω=u'=2 rad/s

α=u''=4 rad/s

r=e^u

So the transverse component of acceleration are given as

a_{\theta}=(ru''+2r'u')\\

Here

r=e^u\\r=e^{\pi/4}\\r=2.1932 m

r'=e^u.u'\\r'=2.1932 \times 2\\r'=4.3864 m

So

a_{\theta}=(ru''+2r'u')\\a_{\theta}=(2.1932\times 4+2\times 4.3864 \times 2)\\a_{\theta}=26.32 m/s\\

The transverse component of acceleration is 26.32 m/s^2

The radial component is given as

a_r=r''-r\theta'^2

Here

r''=e^u.u'^2+e^u u''\\r''=2.1932 \times (2)^2+2.1932\times 4\\r''=17.5456 m

So

a_r=r''-ru'^2\\a_r=17.5456-2.1932\times (2)^2\\a_r=8.7728 m/s^2

The radial component of acceleration is 8.77 m/s^2

6 0
3 years ago
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