A new planet has been discovered that has a mass one-tenth that of Earth and a radius that is one-sixth that of Earth. Determine the free fall acceleration on the surface of this planet. Express your answer in the appropriate mks units.
g=GM/r^2 =g(.1)/(1/6)^2=3.6g
To determine the free fall acceleration on the surface of the planet, we can use the formula for the acceleration due to gravity:
g = G * M / R^2
where g is the acceleration due to gravity, G is the gravitational constant (approximately 6.67 x 10^-11 m^3 kg^-1 s^-2), M is the mass of the planet, and R is the radius of the planet.
Given that the mass of the new planet is one-tenth that of Earth (M = 1/10 * Mass of Earth) and the radius is one-sixth that of Earth (R = 1/6 * Radius of Earth), we can substitute these values into the equation:
g = G * (1/10 * Mass of Earth) / (1/6 * Radius of Earth)^2
Since the problem does not provide the specific values for the Mass of Earth and the Radius of Earth, we will use standard values:
Mass of Earth = 5.97 x 10^24 kg
Radius of Earth = 6.37 x 10^6 m
Substituting these values into the equation:
g = (6.67 x 10^-11 m^3 kg^-1 s^-2) * (1/10 * 5.97 x 10^24 kg) / (1/6 * 6.37 x 10^6 m)^2
Simplifying the equation:
g = (6.67 x 10^-11 m^3 kg^-1 s^-2) * (0.1 * 5.97 x 10^24 kg) / (1/36 * 6.37^2 x 10^12 m^2)
g = (6.67 x 10^-11 m^3 kg^-1 s^-2) * (0.1 * 5.97 x 10^35 kg) / (3.84 x 10^13 m^2)
g = (0.667 x 5.97 x 10^35) / (38.4 x 10^13) m/s^2
g = (3.99 x 10^35) / (38.4 x 10^13) m/s^2
g = 0.10381 m/s^2
Therefore, the free fall acceleration on the surface of this planet is approximately 0.10381 m/s^2.
To determine the free fall acceleration on the surface of the new planet, we can use the formula for the acceleration due to gravity:
a = G * (M/R^2)
Where:
- a is the acceleration due to gravity
- G is the gravitational constant (approximately 6.67430 x 10^-11 m^3/kg/s^2)
- M is the mass of the planet
- R is the radius of the planet
In this case, we are given that the mass of the new planet (M') is one-tenth that of Earth (M) and the radius of the new planet (R') is one-sixth that of Earth (R).
So, we can substitute the given values into the formula:
a = G * (M'/R'^2)
Since M' = (1/10)M and R' = (1/6)R, we can substitute these values into the formula:
a = G * ( (1/10)M / (1/6)^2 )
Simplifying further:
a = G * ( (1/10)M / (1/36) )
To simplify the expression, we can multiply the numerator and denominator by 36 to get rid of the fraction:
a = G * ( (1/10)M * 36 / 1 )
Simplifying:
a = G * ( 36/10 ) * M
Finally, we can express the answer in appropriate MKS units by substituting the value of G (6.67430 x 10^-11 m^3/kg/s^2):
a = (6.67430 x 10^-11 m^3/kg/s^2) * ( 36/10 ) * M
Therefore, the free fall acceleration on the surface of the new planet is given by this expression.