What is the acceleration due to gravity at a height R/5 from the surface of the Earth (radius R)?
Answers
Given:-
•Height from the surface of earth=R/5
•Radius of the earth=R
To find:-
•Acceleration due to gravity at a height R/5 from the surface of the earth
Solution:-
•Value of acceleration due to gravity at the surface of earth=9.8m/s²
If the value of acceleration due to gravity at a height h above the surface of earth is g',then:-
=>g=GMe/R²e -------(1)
=>g'=GMe/(Re+h)²--------(2)
From,equation (1) and (2)
=>g'=g/(1+h/Re)²
=>g'=9.8/(1+R/5/R)²
=>g'=9.8(1+1/5)²
=>g'=9.8/(36/25)
=>g'=9.8×25/36
=>g'=6.8m/s²
Thus,acceleration due to gravity at a height R/5 from the surface of earth is 6.8m/s².
Explanation:
Answer :-✍️
Acceleration of the body is 0.0042 m/s² .
Explanation :-
From the graph, we have :-
→ Initial velocity of the body (v₁) = 20 m/s
→ Final velocity of the body (v₂) = 80 m/s
→ Time interval (t) = 4 hrs
= 4(3600)
= 14400 s
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Now, we know that :-
Acceleration = Change in Velocity/Time
⇒ a = ∆v/t
⇒ a = (v₂ - v₁)/t
⇒ a = (80 - 20)/14400
⇒ a = 60/14400
⇒ a = 6/1440
⇒ a = 0.0042 m/s²
Some Extra Information :-
The 3 equations of motion, for a body moving with uniform acceleration are :-
• v = u + at
• s = ut + ½at²
• v² - u² = 2as
Given:-
•Height from the surface of earth=R/5
•Radius of the earth=R
To find:-
•Acceleration due to gravity at a height R/5 from the surface of the earth
Solution:-
•Value of acceleration due to gravity at the surface of earth=9.8m/s²
If the value of acceleration due to gravity at a height h above the surface of earth is g',then:-
=>g=GMe/R²e -------(1)
=>g'=GMe/(Re+h)²--------(2)
From,equation (1) and (2)
=>g'=g/(1+h/Re)²
=>g'=9.8/(1+R/5/R)²
=>g'=9.8(1+1/5)²
=>g'=9.8/(36/25)
=>g'=9.8×25/36
=>g'=6.8m/s²
Thus,acceleration due to gravity at a height R/5 from the surface of earth is 6.8m/s².