Question

Kepler-62e is an exoplanet that orbits within the habitable zone around its parent star. The planet has a mass that is 3.57 times larger than Earth's and a radius that is 1.61 times larger than Earth's. Kepler-62e is an exoplanet that orbits within the habitable zone around its parent star. The planet has a mass that is 3.57 times larger than Earth's and a radius that is 1.61 times larger than Earth's. Kepler-62e is an exoplanet that orbits within the habitable zone around its parent star. The planet has a mass that is 3.57 times larger than Earth's and a radius that is 1.61 times larger than Earth's. Calculate the acceleration of gravity on the surface of Kepler-62e.

Answer 1

g' = 13.5 m/s²

Step-by-step explanation:

The acceleration due to gravity on surface of earth is given by the formula:

g = GMe/Re² --------------- euation 1

where,

g = acceleration due to gravity on surface of earth

G = Universal Gravitational Constant

Me = Mass of Earth

Re = Radius of Earth

Now, the the acceleration due to gravity on the surface of Kepler-62e is:

g' = GM'/R'² --------------- euation 1

where,

g' = acceleration due to gravity on surface of Kepler-62e

G = Universal Gravitational Constant

M' = Mass of Kepler-62e = 3.57 Me

R' = Radius of Kepler-62e = 1.61 Re

Therefore,

g' = G(3.57 Me)/(1.61 Re)²

g' = 1.38 GMe/Re²

using equation 1:

g' = 1.38 g

where,

g = 9.8 m/s²

Therefore,

g' = 1.38(9.8 m/s²)

g' = 13.5 m/s²