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attashe74 [19]
3 years ago
11

g A meteoroid is in a circular orbit 600 km above the surface of a distant planet. The planet has the same mass as Earth but has

a radius that is 90 % of Earth's (where Earth's radius is approximately 6370 km ). What is the gravitational acceleration of the meteoroid
Physics
1 answer:
Nikitich [7]3 years ago
3 0

Answer:

Explanation:

gravitational acceleration of meteoroid

=  GM / R²

M is mass of planet , R is radius of orbit of meteoroid from the Centre of the planet .

R = (.9 x 6370 + 600 )x 10³ m

= 6333 x 10³ m

M , mass of the planet = 5.97 x 10²⁴ kg .

gravitational acceleration of meteoroid

=  GM / R²

=  (6.67 x 10⁻¹¹ x 5.97 x 10²⁴ kg  / (6333 x 10³ m)²

9.92m/s²

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-\frac{8\pi}{3}rad/s^2

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Answer:

a) the reflective surface has twice the energy transfer

b) A = 1.3 10²⁷ km²

c) the energy emitted by the sun is distributed in a sphere that depends on the square of the distance, and the gravitational force depends on the square of the distance

Explanation:

a) The pressure exerted on the candle is related to the variation of the momentum

          P = \frac{1}{c} \ \frac{dp}{dt}

in a case of absorption (inelastic shock) all the energy is absorbed therefore the pressure is

          P = \frac{1}{c} \ \frac{dp}{dt}

in the case of reflection (elastic shock) an energy is absorbed by absorbing the light and then by action and reaction the same energy is absorbed in the reflected light

          P = 2  \frac{1}{c} \ \frac{dp}{dt}

In conclusion, the reflective surface has twice the energy transfer.

             

b) pressure is defined with force per unit area

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           F = P A

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           Fg = G m Ms / r²

let's look for the case that the two forces are equal

           F = Fg

           P A_sail = G m Ms = r²

suppose a fully reflective sail

           2 \frac{S}{c}  \ A_{sail} = G \frac{m M_s}{r^2}

           

The pointing vector is the power delivered per unit area

           S = I = P / A

where A is the area of ​​the sphere where the is distributed by the sun

            A = 4π r²

we substitute

            \frac{2P}{c} \ \frac{A_{sail}}{4 \pi   r^2} = G \frac{m M_s}{r^2}

            \frac{1}{2 \pi \ c } A_{sail} = G m M_s

            A = G  m M_s 2π c

let's calculate

            A = 6.67 10⁻¹¹  10000 2 10³⁰ 2π  3 10⁸

            A = 1,257 10³³ m²

let's reduce to km²

            A = 1.3 10³³ m² (1km / 10³ m) ²

            A = 1.3 10²⁷ km²

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