At what height from the surface of earth will the value of g be reduced by `36%` from the value on the surface? Take radius of earth `R = 6400 km`.
A
3.2×10^6 m
B
1.6×10^6 m
C
2.5×10^6 m
D
4.6×10^6 m
Text Solution
Verified by Experts
The correct Answer is:
B
Since the acceleration due to gravity reduces by `36%` the value of acceleration due to gravity there is `100 -36=64%`. It means `g'=64/100g.` If `h` is the height of location above the surface of earth then `g'=gR^(2)/((R+h)^2)` or `64/100g=gR^(2)/((R+h^(2))^(2))` `8/10=R/(R+h)` or `8R+8h=10R` `impliesh=(2R)/8=R/4=(6.4xx10^(6))/4=1.6xx10^(6)m`
Topper's Solved these Questions
GRAVITATION
CENGAGE PHYSICS|Exercise Exercise 6.1|13 Videos
GRAVITATION
CENGAGE PHYSICS|Exercise Exercise 6.2|15 Videos
FLUID MECHANICS
CENGAGE PHYSICS|Exercise INTEGER_TYPE|1 Videos
KINEMATICS-1
CENGAGE PHYSICS|Exercise Integer|9 Videos
Similar Questions
Explore conceptually related problems
At what height from the surface of earth will the value of g becomes 40% from the value at the surface of earth. Take radius of the earth = 6.4 xx 10^(6) m .
At what height above the earth's surface does the value of g becomes 36% of the value at the surface of earth ?
At what height from Earth's surface the acceleration due to gravity becomes 20% of its value on the surface of earth. [Radius of earth = 6,400 km.]
How much above the surface of earth does the accelration due to gravity reduces by 64% of its value on the earth. Radius of earth = 6400 km .
At what depth from the surface of earth, the value of acceleration due to gravity is reduced by 40% of its value on the surface of earth. Radius of the earth = 6.4 xx 10^(6) m .
How high above the surface of the Earth does the acceleration due to gravity reduce by 36% of its value on the surface of the Earth ?
At what height above the surface of earth the value of "g" decreases by 2 % [ radius of the earth is 6400 km ]
