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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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Let's begin by stating clear that motiont is the change of position of a body at a certain time. So, during this motion, the balloon will have a trajectory and a displacement, being both different:

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But, if we talk about displacement, we have to draw a straight line between the initial position of the balloon (point 0) to its final position (point 90 m).  Being its displacement 95 m.

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A 5.00μF parallel-plate capacitor is connected to a 12.0 V battery. After the capacitor is fully charged, the battery is disconn
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(a) 12.0 V

In this problem, the capacitor is connected to the 12.0 V, until it is fully charged. Considering the capacity of the capacitor, C=5.00 \mu F, the charged stored on the capacitor at the end of the process is

Q=CV=(5.00 \mu F)(12.0 V)=60 \mu C

When the battery is disconnected, the charge on the capacitor remains unchanged. But the capacitance, C, also remains unchanged, since it only depends on the properties of the capacitor (area and distance between the plates), which do not change. Therefore, given the relationship

V=\frac{Q}{C}

and since neither Q nor C change, the voltage V remains the same, 12.0 V.

(b) (i) 24.0 V

In this case, the plate separation is doubled. Let's remind the formula for the capacitance of a parallel-plate capacitor:

C=\frac{\epsilon_0 \epsilon_r A}{d}

where:

\epsilon_0 is the permittivity of free space

\epsilon_r is the relative permittivity of the material inside the capacitor

A is the area of the plates

d is the separation between the plates

As we said, in this case the plate separation is doubled: d'=2d. This means that the capacitance is halved: C'=\frac{C}{2}. The new voltage across the plate is given by

V'=\frac{Q}{C'}

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A=\pi r^2

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C'=\frac{\epsilon_0 \epsilon_r A'}{d}=4\frac{\epsilon_0 \epsilon_r A}{d}=4C

So, the capacitance has increased by a factor 4; therefore, the new voltage is

V'=\frac{Q}{C'}=\frac{Q}{4C}=\frac{1}{4} \frac{Q}{C}=\frac{V}{4}

which means

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