The radius of the moon is 27%of the Earth's radius and its mass is 1.2% of the earth's mass. Find the acceleration due gravity on the surface of moon. How much will a 424N body weight there? Physics class 11th problems solution Physics class 11th numericals solution First year physics problems solution First year physics numericals solution Gravitation problems solution
00:00Question is, the radius of the moon is 27% of the Earth's radius and its mass is 1.2%
00:16of the Earth's mass.
00:19a Find the acceleration due to gravity on the surface of the moon.
00:24b How much will a 424 Newton body weigh there?
00:32So we have to find the value of g on the surface of moon and weight of 424 Newton body on the
00:42surface of moon.
00:45So first of all, we will make a data is the radius of moon is 27% of the radius of Earth.
01:06So radius of moon will be equal to 27 divided by 100 because the value of percentage is
01:151 upon 100 into the radius of Earth.
01:20So it will be 0.27 radius of the Earth means radius of moon is equal to 0.27 radius of the
01:39Earth.
01:43And mass of the moon is equal to 1.2% of the mass of Earth.
01:55So mass of moon will be equal to 1.2 divided by 100 because the value of percentage is
02:031 upon 100 mass of Earth which will be equal to 0.012 mass of Earth.
02:14So mass of moon will be equal to 0.012 mass of Earth.
02:28So we have to find the value of g on the surface of moon and weights of 424 body on the surface
02:39of moon is weight on the Earth is obtained is 424 Newton.
02:59So for solution, first of all, we will find the value of g on the surface of moon is on the
03:13surface of Earth.
03:24The g is calculated by the formula g is equal to g m e upon r e square.
03:35So on the surface of moon, so on the surface of moon, this formula will become as g on moon
03:48is equal to g into mass of moon divided by radius of moon square.
04:03g on moon will be equal to g into mass of moon which is obtained as 0.012 mass of Earth divided
04:18by radius of moon which is obtained as 0.27 radius of Earth whole square.
04:31So g on moon will be obtained as 0.012 enter g m e upon the square of 0.27 will be 0.0729.
04:38the square of r e will be r e square.
04:45So g on moon will be obtained as 0.012 divided by 0.012 upon the square of 0.27 will be 0.0729.
04:52So g on moon will be obtained as 0.012 divided by 0.0729.
04:59g on moon will be obtained as 0.012 divided by 0.0729 into g m e upon r e square.
05:06So g on moon will be obtained as 0.012 divided by 0.012 divided by 0.0729 into g m e upon r e square.
05:13So g on moon will be obtained as 0.012 divided by 0.0729 into g m e upon r e square.
05:30So after simplifying this we will get the value of g on moon as 0.164 into s g m e upon r e square
05:51is equal to g. So this is our first value or further we can simplify it as g on the surface of earth is 9.8 meter per second square.
06:11So g on moon will be obtained as 0.164 into 9.8.
06:20So g on moon will be obtained as 1.613 meter per second square.
06:31So this is our final required answer is the value of g on the surface of earth is 9.8 meter per second square.
06:43And the value of g on the surface of moon is 1.613 meter per second square.
06:51Now weight on moon which is equal to mass and the value of g on moon.
07:05So first of all we will find the mass is mass is equal to weight upon g.
07:16As weight is obtained as 424 Newton on the surface of earth.
07:23So 424 divided by g which is 9.8 on the surface of earth.
07:30So mass will be obtained as 43.265 kilogram.
07:40Now weight on the moon will be obtained as mass is 43.265 into the value of g on the surface of moon is obtained as 1.613 meter per second square.
08:05So weight on the moon will be obtained as weight on the moon will be obtained as when we will multiply these values we will get the answer as 69.786 Newton.
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