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IceJOKER [234]
3 years ago
11

If I could snap my fingers and move Earth to an orbital distance at 9 AU, what would happen to the strength of the Sun's gravita

tional force on Earth?
Physics
1 answer:
Vitek1552 [10]3 years ago
6 0

Answer:

    \frac{F}{F_o} = 1.2 10⁻²

force of attraction of the sun decreases by a factor of 1.2 10⁻² times

Explanation:

The force of gravity is given by the universal law of attraction

          F = G m Ms / r²

where m is the mass of the Earth, Ms the mass of the Sun and r the distance from the Earth to the Sun

for the initial orbit

          F₀ = G m Ms / R₀²

for one orbit

          R = 9R₀

         F = G \frac{m M_s}{ ((9R_o)^2}

         F = G \frac{m M_s}{81 \  R_o^2}

we substitute

         F = F₀ / 81

         \frac{F}{F_o} = 1.2 10⁻²

therefore the force of attraction of the sun decreases by a factor of 1.2 10⁻² times

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A speaker vibrates at a frequency of 200 hz what is its period
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In this problem, m_1 = 175 kg, m_2 =15 kg and r=2 m, therefore the gravitational force between the two objects is
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Explanation:

We can solve this problem by using the law of conservation of energy.

In absence of frictional effect, the mechanical energy of the apple must be conserved during the fall. So we can write:

U_i +K_i = U_f + K_f

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U_i is the initial potential energy, at the top

K_i is the initial kinetic energy, at the top

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By explicing the potential energy, we can rewrite the equation as:

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Therefore we can solve the equation for K_f, the final kinetic energy of the ball:

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