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ANTONII [103]
2 years ago
8

For a satellite of mass mS in a circular orbit of radius rS around the Earth, determine its kinetic energy, K . Express your ans

wer in terms of some or all of the variables mS, rS, and constants G and ME.
Physics
1 answer:
agasfer [191]2 years ago
8 0

Since it is asking you to find the kinetic energy in relation to the mass, radius, mechanical energy (total energy), and constants, you will need to setup an equation first to "find" the Mechanical Energy, so that we can then solve for the kinetic energy, as from my experience with high school physics, there is only the graviational potential energy equation and force in relation to celestial bodies.

Knowing the ME is the total energy, we add up the energies of the system. Since it is being influenced by the Earth, as per the problem stating the satellite has circular orbit around the Earth, we know there is gravitational potential. Since it is orbiting, we can assume some type of velocity. Nothing else that we need to worry about should be occuring at this level of physics, leaving you with

ME= Ug+K

from here we solve for K, as plugging in could get confusing and messy at the moment.

ME-Ug=K

now using the equations presumably given in class, if not then using this equation, we can find the Ug

Ug=(-(Gm*M)/r)     note that M is the mass of the Earth and m is the satellite

this should give us

ME-(-(GmM)/r)=K

since there is a negative being subracted, we can change that to

ME+(GmM)/r=K

I believe this should be fine, as the Earth's mass is constant, but if not, then all you need to figure left is how to get rid of the M in the equation, as the rest of the terms and constants are for sure within the requirements.

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Answer:

a) t=2s

b) h_{max}=1024ft

c) v_{y}=-256ft/s

Explanation:

From the exercise we know the initial velocity of the projectile and its initial height

v_{y}=64ft/s\\h_{o}=960ft\\g=-32ft/s^2

To find what time does it take to reach maximum height we need to find how high will it go

b) We can calculate its initial height using the following formula

Knowing that its velocity is zero at its maximum height

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

0=(64ft/s)^2-2(32ft/s^2)(y-960ft)

y=\frac{-(64ft/s)^2-2(32ft/s^2)(960ft)}{-2(32ft/s^2)}=1024ft

So, the projectile goes 1024 ft high

a) From the equation of height we calculate how long does it take to reach maximum point

h=-16t^2+64t+960

1024=-16t^2+64t+960

0=-16t^2+64t-64

Solving the quadratic equation

t=\frac{-b±\sqrt{b^{2}-4ac}}{2a}

a=-16\\b=64\\c=-64

t=2s

So, the projectile reach maximum point at t=2s

c) We can calculate the final velocity by using the following formula:

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

v_{y}=±\sqrt{(64ft/s)^{2}-2(32ft/s^2)(-960ft)}=±256ft/s

Since the projectile is going down the velocity at the instant it reaches the ground is:

v=-256ft/s

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Answer:

The  electric field  is 35\cos(90\pi t)\ mV/m

Explanation:

Given that,

Radius = 2.00 cm

Number of turns per unit length n= 1.65\times10^{3}

Current I = 6.00\sin 90\pi t

We need to calculate the induced emf

\epsilon =\mu_{0}nA\dfrac{dI}{dt}

Where, n = number of turns per unit length

A = area of cross section

\dfrac{dI}{dt}=rate of current

Formula of electric field is defined as,

E=\dfrac{\epsilon}{2\pi r}

Where, r = radius

Put the value of emf in equation (I)

E=\dfrac{\mu_{0}nA\dfrac{dI}{dt}}{2\pi r}....(II)

We need to calculate the rate of current

I=6.00\sin 90\pi t....(III)

On differentiating equation (III)

\dfrac{dI}{dt}=90\pi\times6.00\cos(90\pi t)

Now, put the value of rate of current in equation (II)

E=\dfrac{4\pi\times10^{-7}\times1.65\times10^{3}\times\pi\times(2.00\times10^{-2})^2\times90\pi\times6.00\cos(90\pi t)}{2\pi\times 2.00\times10^{-2}}

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Answer:

Vc = 2.41 v

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