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3 years ago

Gravitational Acceleration inside a Planet

Physics

1 answer:

vazorg [7]3 years ago

3 0

Answer:

PART A

g(R) = $\frac{G(rho)(4)\pi R}{3}$ ;

PART B

g(R) = g(Rp) × $\frac{R}{Rp}$ .

Explanation:

Given density of planet = rho.

The planet's radius = Rp.

An object is located a distance R from the center of the planet,

where R< Rp.

The gravitational fore between two point masses m₁ and m₂ is,

F = $\frac{Gm_{1}m_{2}}{r^{2} }$ ; G= universal gravitational constant

r = distance between the masses.

for mass m₂ , F= m₂ g; where g = acceleration due to gravity;

so, g = $\frac{F}{m_{2} }$ = $\frac{Gm_{1}}{r^{2} }$ ;

From figure, only inside part of the planet exerts force and which can be treated as a point mass.

so, g = $\frac{Gm_{R}}{R^{2} }$

where $m_{R}$ = mass of the planet with radius R.

⇒ $m_{R}$ = rho × $\frac{4}{3}$ $×\pi$ ×R³

⇒ g(R) = $\frac{G(rho)(4)\pi R}{3}$ → PART A

PART B

At the surface g(Rp) = $\frac{G(rho)(4)\pi Rp}{3}$

⇒ g(R) = g(Rp) × $\frac{R}{Rp}$

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1 year ago

Lagrangian mechanics. Determine the equations of motion for a particle of mass m constrained to move on the surface of a cone in

maria [59]

Answer:

Explanation:

Hi!

In order to obtain the Lagrangian of the system we must first write the Kinetic and Potential Energies. Lets orient our axes such that the axis of the cone coincide with the z axis. In cilindrical coordinates we have

$v^{2} = \frac{dr}{dt}^{2} +r^{2} \frac{d\theta }{dt} ^{2} +\frac{dz}{dt} ^{2}$ - (1)

But, since the particle is constrained to move on the surface of the cilinder, we have the following relation between r and z:

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and:

$\frac{dz}{dt} = \frac{dr}{dt} cot(45)$

replacing (2) in (1) we obtain:

$v^{2} = \frac{dr}{dt}^{2} (1+cot(45))+r^{2}\frac{d\theta }{dt} ^{2}$ - (3)

Now the kinetic energy is given as:

$T = \frac{1}{2}m(\frac{dr}{dt}^{2} (1+cot(45))+r^{2}\frac{d\theta }{dt} ^{2})$ - (4)

And the potential energy is given by:

$V = -mgz = -mgr cot(45)$

So the Langrangian is given by:

$L = T - V= \frac{1}{2}m(\frac{dr}{dt}^{2}(1+cot(45)+r^{2})\frac{d\theta }{dt} ^{2}) + mgr cot(45)$

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3 years ago

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Where we have recognized the expression for the initial force between the particles, and replaced it with $F_i$ to make the new relation obvious.

Therefore, the new gravitational force between the two masses is $\frac{3}{2}$ of the original force.

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3 years ago

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