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grin007 [14]
3 years ago
8

If it takes a planet 2.8 × 108 s to orbit a star with a mass of 6.2 × 1030 kg, what is the average distance between the planet a

nd the star?
Physics
1 answer:
Levart [38]3 years ago
5 0

Answer:

9.36*10^11 m

Explanation

Orbital velocity  v=√{(G*M)/R},

G = gravitational constant =6.67*10^-11 m³ kg⁻¹ s⁻²,

M = mass of the star

R =distance from the planet to the star.

v=ωR, with ω as the angular velocity and R the radius

ωR=√{(G*M)/R},

ω=2π/T,

T = orbital period of the planet

To get R we write the formula by making R the subject of the equation

(2π/T)*R=√{(G*M)/R}

{(2π/T)*R}²=[√{(G*M)/R}]²,

(4π²/T²)*R²=(G*M)/R,

(4π²/T²)*R³=G*M,

R³=(G*M*T²)/4π²,

R=∛{(G*M*T²)/4π²},

Substitute values

R=9.36*10^11 m

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The emerging velocity of the bullet is <u>71 m/s.</u>

The bullet of mass <em>m</em> moving with a velocity <em>u</em>  has kinetic energy. When it pierces the block of wood, the block exerts a force of friction on the bullet. As the bullet passes through the block, work is done against the resistive forces exerted on the bullet by the block. This results in the reduction of the bullet's kinetic energy. The bullet has a speed <em>v</em> when it emerges from the block.

If the block exerts a resistive force <em>F</em> on the bullet and the thickness of the block is <em>x</em> then, the work done by the resistive force is given by,

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Divide equation (2) by equation (1) and simplify for v<em>₁.</em>

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Thus the speed of the bullet is 71 m/s


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