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Doss [256]
2 years ago
13

Suppose we have two planets with the same mass, but the radius of the second one is twice the size of the first one. How does th

e free-fall acceleration on the second planet compare to the first planet
Physics
1 answer:
bogdanovich [222]2 years ago
6 0

The free-fall acceleration on the second planet is one-fourth the value of the first planet.

Calculation:

Consider the mass of planet A to be, M

               the mass of planet B to be, Mₓ = M

               the radius of planet A to be, R₁

               the radius of planet B to be, R₂

The acceleration due to gravity on planet A's surface is given as:

g = GM/R₁²      - (1)

Similarly, the acceleration due to gravity on planet B's surface is given as:

g' = GM/R₂²                           [where, R₂ = 2R₁]

   = GM/4R₁²    -(2)

From equation 1 & 2, we get:

g/g' = GM/R₁² ÷ GM/4R₁²

g/g' = 4/1

Thus we get,

g' = 1/4 g

Therefore, the free-fall acceleration on the second planet is one-fourth the value of the first planet.

Learn more about free-fall here:

<u>brainly.com/question/13299152</u>

#SPJ4

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The shape is connected in parallel so;

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\frac{1}{R}  =  \frac{1}{R1} +  \frac{1}{R2}   \\  \frac{1}{R}  =  \frac{1}{2}  +  \frac{1}{3}  \\  \frac{1}{R}  =  \frac{3 + 2}{6}  =  \frac{5}{6}  \\ R =  \frac{6}{5}  = 1.2 \:  \: ohm

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\frac{1}{R}  =  \frac{1}{R1} +  \frac{1}{R2}   \\  \frac{1}{R}  =  \frac{1}{8}  +  \frac{1}{10}  \\  \frac{1}{R}  =  \frac{5 + 4}{40}  =  \frac{9}{40}  \\ R =  \frac{40}{9}  = 4.4 \:  \: ohm

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