Asked by UK
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.
Answers
Answered by
bobpursley
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
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
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