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Anettt [7]
2 years ago
11

Two identical asteroids travel side by side while touching one another. If the asteroids are composed of homogeneous pure iron a

nd are spherical, what diameter in feet must they have for their mutual gravitational attraction to be 1 lb?

Physics
1 answer:
victus00 [196]2 years ago
6 0

Answer:

diameter = 21.81 ft

Explanation:

The gravitational force equation is:

  1. F=\frac{GMm}{R^{2} }

Where:

  • F => Gravitational force or force of attraction between two masses
  • M => Mass of asteroid 1
  • m => Mass of asteroid 2
  • R => Distance between asteroids 1 and 2 (from center of gravity)

We also know that the asteroids are identical so their masses are identical:

  • M=m

Since R is the distance between centers of the two asteroids and their diameters are identical (see attachment), we can conclude that:

  • R=d=2r

We don´t know the mass of the asteroids but we know they are composed of pure iron, so we can relate their masses to their density:

  • m=ρV

This is going to be helpful because the volume of a sphere is:

  • \frac{4}{3}\pi r^{3}

And know we can write our original force of gravity equation in terms of the radius of the asteroids:

  • F=\frac{GMm}{R^{2} } =\frac{Gmm}{(2r)^{2} } =\frac{Gm^{2} }{4r^{2} }
  • F=\frac{G ( \frac{4}{3}\pi r^{3}ρ)^{2} }{4r^{2} }
  • F= \frac{G(16)\pi ^{2} r^{6} ρ^{2}}{(9)(4)r^{2} } =\frac{G(16)\pi ^{2} r^{4}ρ^{2}  }{36}

Now let´s plug in the values we know:

  1. F = 1 lb     mutual gravitational attraction force
  2. G = 6.67(10)^{-11}     gravitational constant
  3. ρ_{iron} =491.5 \frac{lb}{ft^{3} }

  • 1= \frac{6.67(10)^{-11} \pi ^{2} r^{4} (491.5)^{2}}{36}

Solve for r and multiply by 2 because 2r = diameter

  • d=2\sqrt[4]{\frac{1}{7.07(10)^{-5} } }

Result is d = 21.81 Feet

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The total resistance in the circuit, R is 4 Ω .

The meters connected to the 12 Ω resistance are:

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  2. Ammeter - connected in series

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Equivalent resistance = 4 + 2 = 6 Ω

The 4 Ω and the 2 Ω resistances are then connected in parallel with the 12 Ω resistance.

Total resistance, R is calculated as follows:

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The meters connected to the 12 Ω resistance are:

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