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GrogVix [38]
2 years ago
5

Derive an expression for the gravitational potential energy of a system consisting of Earth and a brick of mass m placed at Eart

h's center. Take the potential energy for the system with the brick placed at infinity to be zero. Express your answer in terms of or all of the variables m, mass of Earth mE, its radius RE, and gravitational constant G.
Physics
1 answer:
Arlecino [84]2 years ago
5 0

Answer:

The gravitational potential energy of a system is -3/2 (GmE)(m)/RE

Explanation:

Given

mE = Mass of Earth

RE = Radius of Earth

G = Gravitational Constant

Let p = The mass density of the earth is

p = M/(4/3πRE³)

p = 3M/4πRE³

Taking for instance,a very thin spherical shell in the earth;

Let r = radius

dr = thickness

Its volume is given by;

dV = 4πr²dr

Since mass = density* volume;

It's mass would be

dm = p * 4πr²dr

The gravitational potential at the center due would equal;

dV = -Gdm/r

Substitute (p * 4πr²dr) for dm

dV = -G(p * 4πr²dr)/r

dV = -G(p * 4πrdr)

The gravitational potential at the center of the earth would equal;

V = ∫dV

V = ∫ -G(p * 4πrdr) {RE,0}

V = -4πGp∫rdr {RE,0}

V = -4πGp (r²/2) {RE,0}

V = -4πGp{RE²/2)

V = -4Gπ * 3M/4πRE³ * RE²/2

V = -3/2 GmE/RE

The gravitational potential energy of the system of the earth and the brick at the center equals

U = Vm

U = -3/2 GmE/RE * m

U = -3/2 (GmE)(m)/RE

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Answer:

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At the point 1

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Z₂= 0 m/s

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Now by putting the values

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3-1.5=\dfrac{V_2^2}{2\times 9.81}

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V₂= 5.42 m/s

The initial velocity of the water from the tank is 5.42 m/s

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3 years ago
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A 900-kg car traveling east at 15.0 m/s collides with a 750-kg car traveling north at 20.0 m/s. The cars stick together. Assume
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Explanation:

It is given that,

Mass of the car 1, m_1=900\ kg

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Initial speed of car 2, u_1=20j\ m/s (north)

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m_1u_1+m_2u_2=(m_1+m_2)V

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tan\theta=\dfrac{9.09}{8.18}

\theta=48.01^{\circ}

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