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4 years ago

If the moon orbited twice as far from earth how far would it "fall" each second?

1 answer:

5 0

I can't imagine that this is going to do you much good, but
I'm sure going to enjoy solving it.
-------------------------
Skip this whole first section.
It was an attempt to master a bunch of trees, while
the forest was right there in front of me all the time.
Drop down below the double line.
-------------------------

Kepler's 3rd law says:

 (square of the orbital period) / (cube of the orbital radius) = constant

 T₀² = K R₀³

I put the zero subscripts in there, because you doubled 'R'
and I need to know how that affected 'T'.

 new-T² = K(2 R₀)³

 new-T² = 8 K (R₀)³ = 8 old-T₀²

 new-T = √8 old-T <=== that's what I was after

I just teased out the Moon's new orbital period if it's distance were doubled.
Instead of 1 month, it's now √8 months.

To put a somewhat sharper point on it, the moon's period of revolution
changes from 27.322 days to 27.322√8 = 77.278 days (rounded) .

Using 385,000 km for the moon's current average distance, the current orbital speed is
 (2π x 385,000 km) / (27.322 days) = 1,024.7 m/s
(One online source says 1.023 km, so we're not doing too badly so far.)

================================================

I'm such a dummy. I don't need to go through all of that.

If the moon were twice as far from Earth as it really is, then it would
average 770,000 km instead of the present 385,000 km.

That's 120.86 times the Earth's radius of 6,371 km.

So the acceleration of gravity out there would be

 (1 / 120.86)² of the (9.807 m/s²) that it is here on the surface.

 new-G = 0.000671 m/s²

Distance a dropped object falls = 1/2 g t²

 In the first second, that's 1/2 g (1)² = 1/2 g

For an orbiting object, every second is the "first"second, because ...
as we often explain orbital motion qualitatively ... the Earth "falls away"
just as fast as the curved orbit falls.

Distance an object falls in the 1st second =

 1/2 G = 0.000336 m/s = 0.336 millimeter per second

I estimate the probability of a mistake somewhere during this process
at approx 99.99% . But I don't have anything better right now, and I've
wasted too much time on it already, so I'll stick with it.

You might be interested in

A 15 kg cart is pushed on a frictionless surface from rest horizontally by a 30 N force. What is the cart's acceleration?

Rainbow [258]

Answer:

a. The cart's acceleration is 2 m/s^2

b. The cart will travel 100 m

c. The speed is 20 m/s

Explanation:

a. The acceleration of the cart can be calculated using Newton's second law:

F = m.a

Solving for a:

$\displaystyle a=\frac{F}{m}$

The cart has a mass of m=15 Kg and is applied a net force of F=30 N, thus:

$\displaystyle a=\frac{30}{15}$

$a=2\ m/s^2$

Now we use kinematics to find the distance and speed:

$\displaystyle x = v_o.t+\frac{at^2}{2}$

The cart starts from rest (vo=0). The distance traveled in t=10 seconds is:

$\displaystyle x = 0*10+\frac{2*10^2}{2}$

$x = 100\ m$

The cart will travel 100 m

The final speed is calculated by:

$v_f=0+2*10=20\ m/s$

The speed is 20 m/s

5 0

3 years ago

Consider a motor that exerts a constant torque of 25.0 nâ‹…m to a horizontal platform whose moment of inertia is 50.0 kgâ‹…m2. A

Crazy boy [7]

Answer:

W = 1884J

Explanation:

This question is incomplete. The original question was:

Consider a motor that exerts a constant torque of 25.0 N.m to a horizontal platform whose moment of inertia is 50.0kg.m^2 . Assume that the platform is initially at rest and the torque is applied for 12.0rotations . Neglect friction.

How much work W does the motor do on the platform during this process? Enter your answer in joules to four significant figures.

The amount of work done by the motor is given by:

$W=\Delta K$

$W= 1/2*I*\omega f^2-1/2*I*\omega o^2$

Where I = 50kg.m^2 and ωo = rad/s. We need to calculate ωf.

By using kinematics:

$\omega f^2=\omega o^2+2*\alpha*\theta$

But we don't have the acceleration yet. So, we have to calculate it by making a sum of torque:

$\tau=I*\alpha$

$\alpha=\tau/I$ => $\alpha = 0.5rad/s^2$

Now we can calculate the final velocity:

$\omega f = 8.68rad/s$

Finally, we calculate the total work:

$W= 1/2*I*\omega f^2 = 1883.56J$

Since the question asked to "Enter your answer in joules to four significant figures.":

W = 1884J

3 0

3 years ago

A muon has a rest mass energy of 105.7 MeV, and it decays into an electron and a massless particle. If all the lost mass is conv

sergeinik [125]

Answer:

The electron’s velocity is 0.9999 c m/s.

Explanation:

Given that,

Rest mass energy of muon = 105.7 MeV

We know the rest mass of electron = 0.511 Mev

We need to calculate the value of γ

Using formula of energy

$K_{rel}=(\gamma-1)mc^2$

$\dfrac{K_{rel}}{mc^2}=\gamma-1$

Put the value into the formula

$\gamma=\dfrac{105.7}{0.511}+1$

$\gamma=208$

We need to calculate the electron’s velocity

Using formula of velocity

$\gamma=\dfrac{1}{\sqrt{1-(\dfrac{v}{c})^2}}$

$\gamma^2=\dfrac{1}{1-\dfrac{v^2}{c^2}}$

$\gamma^2-\gamma^2\times\dfrac{v^2}{c^2}=1$

$v^2=\dfrac{1-\gamma^2}{-\gamma^2}\times c^2$

Put the value into the formula

$v^2=\dfrac{1-(208)^2}{-208^2}\times c^2$

$v=c\sqrt{\dfrac{1-(208)^2}{-208^2}}$

$v=0.9999 c\ m/s$

Hence, The electron’s velocity is 0.9999 c m/s.

6 0

3 years ago

A heat pump used to heat a house runs about one-third of the time. The house is losing heat at an average rate of 22,000 kJ/h. I

Natali5045456 [20]

The power that heat pump draws when running will be 6.55 kj/kg

A heat pump is a device that uses the refrigeration cycle to transfer thermal energy from the outside to heat a building (or a portion of a structure).

Given a heat pump used to heat a house runs about one-third of the time. The house is losing heat at an average rate of 22,000 kJ/h and if the COP of the heat pump is 2.8

We have to determine the power the heat pump draws when running.

To solve this question we have to assume that the heat pump is at steady state

Let,

Q₁ = 22000 kj/kg

COP = 2.8

Since heat pump used to heat a house runs about one-third of the time.

So,

Q₁ = 3(22000) = 66000 kj/kg

We known the formula for cop of heat pump which is as follow:

COP = Q₁/ω

2.8 = 66000 / ω

ω = 66000 / 2.8

ω = 6.66 kj/kg

Hence the power that heat pump draws when running will be 6.55 kj/kg

Learn more about heat pump here :

brainly.com/question/1042914

#SPJ4

5 0

2 years ago

If a nucleus was as big as a nonpareil, an atom would be ____

Mamont248 [21]

Answer:

small marble

Explanation:

A nonpareil are confectionery tiny ball items that are made up of starch and sugar. It is very small of the size of sugar crystal or sand grains. They were the miniature version of the comfits. They are generally opaque white but bow available in all colors.

In the context, if the nucleus is compared to the size of a nonpareil then its atom would be of the size of small size marble. An atom is bigger in size than that of nucleus as the nucleus is located inside the atoms.

6 0

3 years ago