at what height the acceleration due to gravity is 25% of that at the surface of the earth, in terms of radius of the earth
Answers
Answer:
hi could you tell me if this answer is from the physics paper yesterday, if yes then did you have the paper from before
12700km radius from center of the earth
Given:
Acceleration due to gravity.
Find:
Height at which acceleration due to gravity is 25% of that at the surface of the earth, in terms of radius of the earth.
Formula use:
Force =mass*acceleration =m*g
Force= (GMe*m)/(R*R)
Where: G=gravitational constant
Me=mass of earth
m =mass of body
R=radius of earth
Explanation:
- Force = (GMe*m)/(R*R) =m*g
- Therefore, g=GMe/(R*R)
- In this equation G=gravitational constant and Me=mass of earth will not change. So “g” is inversely proportional to square of radius of earth.
- Normal radius of earth=6350km
- If we reduce “g” by 25% we can increase radius by 2 times. Because in normal case any inversely proportional relation if we reduce one quantity by 0.5 then we increase other by 2 times.
- But in this case it will be square of inversely proportional.
- Hence we can increase radius by 2 times.
- That is 6350*2=12700km
- Hence after 12700km radius from center of the earth value of “g” will be 0.25% of the surface value.
Note: Don't multiply by 4 answer is totally wrong. and they give that option also in answer.
learn more...
1• What is acceleration due to gravity (g) and what is it's SI unit ?
• What will be the value of g on the surface of the Earth ?
https://brainly.in/question/15372835
2.NTSE Sample Question Physics The acceleration due to gravity is zero at
(a) Poles
(b) equator
(c) center of earth
(d) None of these
https://brainly.in/question/207742