The radius of the moon is one eighty-first that of the earth. If the accelaretion due to gravity on the earth is 9.8m/s^2 , what is its value on the moon's surface.
assuming density of each is the same, the mass of the moon...
mass is related to volume = k*4PIr^3
so mass of moon will be (1/81)^3 that of Earth.
"gMoon"= 9.81(1/81)^3/(1/81)^2 = 9.81/81=.12 m/s^2. Now the premise of the problem statement is wrong, the actual radius of the Moon is 1/4 th the radius of Earth, and the density is not the same.
The greater the mass of an object, the stronger the force of gravity. The Moon's mass is about 1.2% of the Earth's mass (1/80th), so the Moon's gravity is much less than the Earth's gravity: 83.3% (or 5/6) less to be exact. Other related facts: the Moon is 1/4 the size of the Earth by diameter, and 1/50 the size by volume. The actual value of the acceleraltion due to gravity on the surface of the moon is ..
"gmoon"= 1.62 m/s^2
To find the acceleration due to gravity on the moon's surface, we can use the formula:
g_moon = (G * M_moon) / (R_moon)^2
Where:
g_moon is the acceleration due to gravity on the moon's surface
G is the gravitational constant
M_moon is the mass of the moon
R_moon is the radius of the moon
Since we are given that the radius of the moon is 1/81st of the Earth's radius, we can write:
R_moon = (1/81) * R_earth
Now, the problem does not provide the mass of the moon. However, we know that the acceleration due to gravity on the Earth is 9.8 m/s^2. Using this information, we can calculate the mass of the moon using the formula:
g_earth = (G * M_earth) / R_earth^2
Rearranging this equation, we have:
M_earth = (g_earth * R_earth^2) / G
Substituting the given values, we can find the mass of the Earth.
Finally, we can substitute the values of M_moon (which we just calculated), R_moon, and G into the formula for the acceleration due to gravity on the moon's surface and solve for g_moon.
I will calculate the result for you.