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VladimirAG [237]
3 years ago
10

. A newly discovered planet has three times the mass and five times the radius of Earth. What is the ratio of the acceleration d

ue to gravity at the surface of the new planet to the acceleration due to gravity at the surface of Earth
Physics
1 answer:
NikAS [45]3 years ago
3 0

Answer:

0.12

Explanation:

The acceleration due to gravity of a planet with mass M and radius R is given as:

g = (G*M) / R²

Where G is gravitational constant.

The mass of the planet M = 3 times the mass of earth = 3 * 5.972 * 10^24 kg

The radius of the planet R = 5 times the radius of earth = 5 * 6.371 * 10^6 m

Therefore:

g(planet) = (6.67 * 10^(-11) * 3 * 5.972 * 10^24) / (5 * 6.371 * 10^6)²

g(planet) = 1.18 m/s²

Therefore ratio of acceleration due to gravity on the surface of the planet, g(planet) to acceleration due to gravity on the surface of the planet, g(earth) is:

g(planet)/g(earth) = 1.18/9.8 = 0.12

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Given:

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<u>Therefore the density of liquid soda:</u>

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s=\frac{\rho}{\rho_w}

where:

\rho_w= density of water

s=\frac{995.5}{1000}

s=0.9955

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