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RSB [31]
1 year ago
15

A planet of mass M and radius R has no atmosphere. The escape velocity at its surface is ve. An object of mass m is at rest a di

stance r from the center of the planet, where r>>R. The particle falls to the surface of the planet. The total mechanical energy of the particle at the surface of the planet is closest to
Physics
1 answer:
zubka84 [21]1 year ago
8 0

Answer:

Explanation:

Expression for escape velocity

ve = \sqrt{\frac{2GM}{R} }

ve² R / 2 = GM

M is mass of the planet , R is radius of the planet .

At distance r >> R , potential energy of object

= \frac{-GMm}{r}

Since the object is at rest at that point , kinetic energy  will be zero .

Total mechanical energy  = \frac{-GMm}{r} + 0 = \frac{-GMm}{r}

Putting the value of GM = ve² R / 2

Total mechanical energy  = ve² Rm / 2 r

This mechanical energy will be conserved while falling down on the earth due to law of conservation of mechanical energy  . So at surface of the earth , total mechanical energy

=  ve² Rm / 2 r

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sergeinik [125]

Answer:

mass of the planet: 5.9\,10^{26}\,kg

Explanation:

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m\frac{v^2}{R}=G\frac{M\,m}{R^2}

where we used here the tangential velocity (v) of the moon around the planet. This equation can be further simplified by dividing both sides by "m" and multiplying both sides by the orbital radius R:

m\frac{v^2}{R}=G\frac{M\,m}{R^2}\\v^2=G\frac{M}{R}

Notice that the mass of the moon has actually disappeared from the equation, which tells us that the orbiting velocity and period do not depend on the mass of the moon, but on the mass of the actual planet.

We know the orbital radius R (5.32\,10^5\,km=5.32\,10^8\,m, the value of the Universal Gravitational constant G, and we can estimate the value of the tangential velocity of the moon since we know it period: 36.3 hrs = 388800 seconds.

We know that the moon makes a full circumference (2\,\pi\,R) in 388800 seconds, therefore its tangential velocity is:

v=\frac{2\,\pi\,5.32\,10^8}{388800} \frac{m}{s} \\v=8.6\,10^3\,\frac{m}{s}

where we rounded the velocity to one decimal.

Notice that we have converted all units to the SI system, so when using the formula to solve for the mass of the planet, the answer comes directly in kg.

Now we use this value for the tangential velocity to estimate the mass of the planet in the first equation we made and simplified:

v^2=G\frac{M}{R}\\M=\frac{v^2\,R}{G} \\M=\frac{(8.6\,10^3)^2\,5.32\,10^8}{6.67\,10^{-11}}kg\\M=5.9\,10^{26}\,kg

8 0
2 years ago
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WARRIOR [948]

Answer:

The speed of ejection is 2.06\times 10^{4}\ m/s

Solution:

As per the question:

Magnetic field density, B = 0.4 T

Density of the material in the sunspot, \rho = 3\times 10^{4}\ kg/m^{3}

Now,

To calculate the speed of ejection of the material, v:

The magnetic field energy density is given by:

U_{B} = \frac{B^{2}}{2\mu_{o}}

This energy density equals the kinetic energy supplied by the field.

Thus

KE = U_{B}

\frac{1}{2}mv^{2} = \frac{B^{2}}{2\mu_{o}}

where

m = mass of the sunspot in 1\ m^{3} = 3\times 10^{- 4}\ kg/m^{3}

v = \frac{B}{\sqrt{\mu_{o}m}}

v = \frac{0.4}{\sqrt{4\pi \times 10^{- 7}\times 3\times 10^{- 4}}} = 2.06\times 10^{4}\ m/s

3 0
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Answer:

Explanation:

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7 0
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The electric field near the earth's surface has magnitude of about 150n/c. what is the acceleration experienced by an electron n
qaws [65]
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Answer:

a=2330

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pb=2330m

t=0.223secs

6 0
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