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Anarel [89]
3 years ago
14

Why would the planets move in a straight path if there was no gravitational energy from the sun​

Physics
2 answers:
omeli [17]3 years ago
6 0

Explanation:

The sun's gravitational force is very strong. If it were not, a planet would move in a straight line out into space. The sun's gravity pulls the planet toward the sun, which changes the straight line of direction into a curve. This keeps the planet moving in an orbit around the sun

Lemur [1.5K]3 years ago
5 0

Answer and Explanation:

The planets would move in a straight path since there would be no source that will be acting upon them (if the sun wasn't there).

Gravitational pull is the amount of force that an object emits. Due to the Sun having a larger mass and a larger gravitational pull, the planets orbit around it.  

If, for example, a house was bigger than everything, including planets and stuff, those things would orbit around the house.

A real life example is the moon orbiting Earth. The Earth has a greater magnitude and mass, so it is able to pull the Moon.

Right now the Sun is the external force, as its gravitational pull is moving the planets around. IF the sun was to magically disappear, the planets won't follow an orbit anymore, since there isn't a gravitational pull on them. This leads them to going in a straight line for possibly a couple years.

Eventually, the planets will stop going in a straight line and start to orbit each other, leading to the planets eventually crashing.

External Force = Sun's gravitational pull because of bigger mass

The planets themselves were created after the sun (though some minerals were created before the sun, but that's minerals and not the planet entirely). The sun gives the planet's their motion. The planets, however (though I'm not sure about mercury and Venus) are going super fast, fast enough to where they aren't being pulled into the sun, but not fast enough to go out of orbit.

If the sun goes out magically, the planets, actually yes will continue orbit due to them being in that motion at such a fast rate, but eventually when they are supposed to turn again, (like, to another part of the sun), the would go straight. Eventually, the will get into the orbit of something else and start going around that object's orbit, most likely outcome being that they crash.

The excess force from no gravitational pull will make the planets go in a straight path. Since nothing is acting upon it, the planets would move in a straight line. Imagine if you were on those spinning circle things, and then let go. You would go around it for like a second and then go immediately straight in one direction.

<em><u>#teamtrees #PAW (Plant And Water)</u></em>

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Thomson's experiments accounted for: the mass of the electron the fundamental charge the force on a current-carrying wire the el
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t = \frac{L}{v}

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the electron's charge-to-mass ratio

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3 years ago
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If the radius of an electron's orbit around a nucleus doubles but the wavelength remains unchanged, what happens to the number o
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Number of electron wavelength will Double

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3 0
3 years ago
Consider a turntable to be a circular disk of moment of inertia It rotating at a constant angular velocity ωi around an axis thr
Rom4ik [11]

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The initial angular momentum of the system is

L = ( It ) ( ωi )

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The final angular momentum is

L = ( It + Ir ) ( ωf )

where ωf is the final angular velocity of the system.

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7 0
3 years ago
A pottery wheel, initially at rest, is assumed to be a uniform disk of mass 3.0 kg and radius 20.0 cm. A 25.0 N force is applied
miv72 [106K]

Answer:

a) 83.33 rad/s^2

b) 2500 rad/s

c) 6187.5 rad

Explanation:

Given data:

radius of wheel = 20.0 cm

mass of wheel = 3.0 kg

force = 25 N

a) angular acceleration

we know torque is given as

T = I \times \alpha

F\times R = I \times \alpha

F\times R = \frac{1}{2} mR^2 \times \alpha

solving for angular acceleration \alpha

\alpha = \frac{2F}{mR} = \frac{2*25}{3*0.2} = 83.33 rad/s^2

b)  angular velocity

duartion of force applied = 30.0 sec

\omega_{30} = \omega_0 +alpha t

                      = o + 83.33 *30 = 2500 rad/sec

c) angular displacement

\theta_{15} = = \omega_0 +\frac{1}{2} alpha t_{15}^2 = 0 + \frac{1}{2} 83.33 \times 15^2

\theta_{30} = \omega_0 +\frac{1}{2} alpha t_{30}^2 = 0 + \frac{1}{2} 83.33 \times 30^2

\Delta \theta = \theta_{30} - \theta_{15}

\Delta \theta = 6187.5 rad

3 0
3 years ago
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