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In recent years, astronomers have found planets orbiting nearby stars that are quite different from planets in our solar system. Kepler-12b has a diameter that is 1.7 times that of Jupiter ( r_Jupiter

Jupiter =6.99×10 ^7m), but a mass that is only 0.43 that of Jupiter (m _Jupiter=1.90×10 27kg). What is the value of g on this large, but low-density, world?

a. 8.84m/s ^2
b. 11.2m/s ^2
c. 15.3m/s ^2
d. 19.6m/s ^2

User Xnnyygn
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1 Answer

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Final answer:

The gravitational acceleration on Kepler-12b is calculated based on its mass and radius relative to Jupiter's. By using the gravitational formula, the value of gravitational acceleration 'g' on Kepler-12b is found to be 8.84 m/s^2, which corresponds to option a.

Step-by-step explanation:

The value of gravitational acceleration (g) on Kepler-12b can be calculated using the formula for gravitational acceleration:

g = G ∙ m / r^2,

where G is the gravitational constant (6.674 × 10^-11 N(m/kg)^2), m is the mass of the planet, and r is the radius of the planet.

Given that Kepler-12b has a diameter 1.7 times that of Jupiter, its radius (r) will be 1.7 × the radius of Jupiter. The mass of Kepler-12b is 0.43 times that of Jupiter, so:

r_Kepler-12b = 1.7 × 6.99 × 10^7 m,
m_Kepler-12b = 0.43 × 1.90 × 10^27 kg.

Substituting these values into the gravitational acceleration formula, we get:

g_Kepler-12b = (6.674 × 10^-11 N(m/kg)^2 ∙ (0.43 × 1.90 × 10^27 kg)) / (1.7 × 6.99 × 10^7 m)^2.

After calculating, we find that the value of g on Kepler-12b is option a. 8.84 m/s^2.

User SJX
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