Physics, asked by krishnakrrish4576, 1 day ago

At what height accleration due to gravity decrease by 25%​

Answers

Answered by gudluuu7rk
0

EXPLAINATION .

Hint: Use the formula for acceleration due to gravity at height h from the surface of the earth. Substitute g4

for acceleration due to gravity at height h and solve the equation for h. For the second part, take the ratio of difference in the acceleration due to gravity at the surface of the earth and acceleration due to gravity at given depth to the acceleration due to gravity at the surface of the earth to determine the percentage decrease in the acceleration due to gravity.

Formula used:

gh=g(RR+h)2

Here, gh

is the acceleration due to gravity at height h from the surface of the earth, g is the acceleration due to gravity at the surface of the earth, and R is the radius of the earth.

gd=g(1−hR)

Here, gd

is the acceleration due to gravity at depth h below the earth’s surface.

Complete step by step answer:

(A) We know the formula for variation of acceleration due to gravity at with altitude h.

gh=g(RR+h)2

Here, gh

is the acceleration due to gravity at height h from the surface of the earth, g is the acceleration due to gravity at the surface of the earth, and R is the radius of the earth.

We have given the acceleration due to gravity at height h is 25%. Therefore, gh=25%g=g4

.

Substitute g4

for gh

in the above equation.

g4=g(RR+h)2

⇒14=(RR+h)2

Take the square root of the above equation, we get,

RR+h=12

⇒2R=R+h

⇒h=R

∴h=6400km

Therefore, at height 6400 km, the acceleration due to gravity is 25% of the acceleration due to gravity at the surface of the earth.

(B) We know the formula for the variation of the acceleration due to gravity with depth below the earth’s surface.

gd=g(1−hR)

Here, gd

is the acceleration due to gravity at depth h below the earth’s surface.

Rearrange the above equation as follows,

gd=g−ghR

⇒g−gd=ghR

∴g−gdg=hR

In the above equation, g−gdg

is the decrease in the acceleration due to gravity with respect to the acceleration due to gravity at the surface of the earth.

Since in the section (A), we have calculated h=R

, the above equation becomes,

g−gdg=RR

g−gdg=1

The percentage decrease in the acceleration due to gravity is,

(g−gdg)%=1×100

(g−gdg)%=100%

Thus, the acceleration due to gravity at height equal to radius of the earth, the decrease in the acceleration due to gravity is 100% that is 0m/s2

.

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