Physics class 11th problem solution
First year physics problem solution
Gravitation problems solution
First year physics problem solution
Gravitation problems solution
Category
📚
LearningTranscript
00:00Question says, what is the value of the gravitational acceleration at a distance of 1 Earth's radius
00:15above the Earth's surface to twice Earth's radius above the Earth's surface?
00:24So first of all, we will make a data is g on the surface of Earth is obtained as 9.8 meter per second squares.
00:44In first condition, we have to find the value of the g Earth's radius above the Earth's surface.
00:54So g dash is equal to what?
01:05So take a suppose this is Earth and this is the radius of Earth.
01:17So we have to find the value of g Earth's radius above the Earth's surface means from the surface
01:27of Earth, we will move a distance equal to radius of Earth.
01:34So here we will find g dash and this is Re.
01:46So from the center of Earth, this distance will be equal to r dash which is equal to the twice
01:56of the radius of the Earth and in second condition, we have to find the value of g twice Earth's
02:07radius above the Earth's surface means double the value of the radius from the surface of
02:14Earth's surface.
02:15So here means it's 2r is from the center of Earth's surface.
02:26So this value becomes r double dash which is equal to 3 Re.
02:46So here we will find g double dash which is equal to what?
02:52So r dash is obtained as 2 Re and in second condition we have to find g double dash.
03:07So in this condition r double dash will be equal to 3 Re.
03:17So for solution is on the surface of Earth, the value of g is means the formula of g is g
03:32m e upon r square.
03:36So this is equation number 1 above the surface of Earth, the value of g will be g dash is
03:43equal to g into m e upon r dash square.
03:57So g dash will be equal to g into m e upon r dash square as the value of r dash is 2 Re.
04:13So 2 Re whole square.
04:19So the value of g dash will be equal to g m e upon the square of 2 is 4.
04:27The square of r is Re square.
04:32So g dash will be equal to 1 upon force into g m e upon Re square is g m e upon Re square
04:48is equal to g m e upon r dash is equal to g m e upon g m e upon g m e upon g m e upon g.
04:57So this is our required condition or further we can simplify this as g dash is equal to 1
05:10upon force into the value of g on the surface of Earth is 9.8 m per second square.
05:18So g dash will be equal to g dash will be equal to 9.8 divided by 4.
05:27So it will be 2.45 m per second square.
05:35So this is our required condition.
05:39Now we will find the value of g double dash means twice the radius from the Earth's surface.
05:55So g double dash will be equal to g m e upon r double dash is equal to 3 Re.
06:14So 3 Re whole square.
06:19So g double dash will be equal to g m e upon the square of 3 is 9 and the square of Re is
06:30Re square.
06:32So g double dash will be equal to 1 upon 9 into g m e upon Re square is g m e upon Re square
06:51is equal to g.
06:52So g double dash will be equal to 1 upon 9 into g or further we can simplify this as g double
07:05dash will be equal to 1 upon 9 into the value of g on the surface of Earth is 9.8 m per second
07:11square.
07:12So g double dash will be equal to g double dash will be equal to 9.8 divided by 9.8 divided
07:18by 9 so it will be 1.088 m per second square.
07:28So this is our required condition.