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1 year ago

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The gravitational potential energy of a particle of mass m moving under the influence of a fixed mass M is given by - , where G

is the universal gravitational constant and r is the distance between the masses. What is the total energy of the mass m if it is in a circular orbit about mass M

1 answer:

djverab [1.8K]1 year ago

3 0

-GMm/2r is the total energy of the mass m if it is in a circular orbit about mass M.

Given

A particle of mass m moving under the influence of a fixed mass's M, gravitational potential energy of formula -GMm/r, where r is the separation between the masses and G is the gravitational constant of the universe.

As the Gravity Potential energy of particle = -GMm/r

Total energy of particle = Kinetic energy + Potential Energy

As we know that

Kinetic energy = 1/2mv²

Also, v is equals to square root of GM/r

v = √GM/r

Put the value of v in the formula of kinetic energy

We get,

Kinetic Energy = GMm/2r

Total Energy = GMm/2r + (-GMm/r)

= GMm/2r - GMm/r

= -GMm/2r

Hence, -GMm/2r is the total energy of the mass m if it is in a circular orbit about mass M.

Learn more about Gravitational Potential Energy here brainly.com/question/15896499

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kari74 [83]

Answer:

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Explanation:

Given:

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charge acquired by the bee, $q_2=23\times 10^{-12}\ C$

a.

Electrical field near the earth surface, $E=100\ N.C^{-1}$

Now the electric force on the bee:

we know:

$F=\frac{1}{4\pi.\epsilon_0} \times \frac{q_1.q_2}{r^2}$

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b.

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So,

$F=9.8\times10^{-4}\ N$

$E.q_2=9.8\times10^{-4}\ N$

$E\times 23\times 10^{-12}=9.8\times10^{-4}\ N$

$E=4.2609\times 10^7\ N.C^{-1}$ parallel to the earth surface.

In this case according to the Fleming's left hand rule the direction of movement of bee must be in a direction parallel to the earth surface and perpendicular to the electric field at the same time.

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