Answer :
The acceleration of the of gravity on the surface of the exoplanet
[tex]Kepler-62e[/tex] orbits within the habitable zone around its parent star is [tex]13.5m/s^{2}[/tex]
The following formula determines the acceleration caused by gravity on Earth's surface:
g is given by the following formula:
[tex]g = GM_{e} / R_{e}^{2}[/tex] equation 1
where,
g = acceleration caused by gravity on earth's surface
G stands for the constant of gravitation, or G.
Me = Earth's mass
Re = Earth's Radius
The acceleration caused by gravity at Kepler-62e's (exoplanet) surface is currently:
g' = GM'/R'²
where,
g' is the acceleration brought on by gravity on Kepler-62e's surface.
G stands for the constant of gravitation.
M' = 3.57 Me, is the mass of Kepler-62e.
Radius of Kepler-62e (R') = 1.61 Re
Therefore,
⇒ g' = G(3.57 Me)/(1.61 Re)²
⇒ g' = [tex]1.38 GMe/Re^{2}[/tex]
using equation 1:
[tex]g' = 1.38 g[/tex]
where,
[tex]g = 9.8 m/s^{2}[/tex]
Therefore,
⇒ [tex]g' = 1.38(9.8 m/s^{2} )[/tex]
⇒[tex]g' = 13.5 m/s^{2}[/tex]
Therefore the acceleration produced on the surface of kepler 62e which is an exoplanet is 13.5 m/s².
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