Question

find the ratio of the acceleration due to gravity at a height R/2 from the the surface of the earth to the value at a depth R/2 from the surface of the Earth (R- radius of the earth)

Answer 1

Answer:

Acceleration due to gravity height h,

g

1

=g(1−

R

2h

)

Acceleration due to gravity at depth h,

g

2

=g(1−

R

h

)

∴

g

2

g

1

=

1−h/R

1−2h/R

=(1−

R

2h

)(1−

R

h

)

−1

=(1−

R

h

)

[neglecting higher power of

R

h

]

∴

g

2

g

1

decreases linearly with h.

Answer 2

The ratio of the acceleration due to gravity is 1:9.

Explanation:

Given that,

Height $h=\dfrac{R}{2}$

Depth $h'=\dfrac{R}{2}$

We need to calculate the gravity at height

Using formula of gravity

$g=\dfrac{GmM}{(R+h)^2}$

Put the value into the formula

$g=\dfrac{GmM}{(R+\dfrac{R}{2})^2}$....(I)

We need to calculate the gravity at depth

Using formula of gravity

$g'=\dfrac{GmM}{(R-h)^2}$

Put the value into the formula

$g'=\dfrac{GmM}{(R-\dfrac{R}{2})^2}$....(II)

We need to calculate the ratio of the acceleration due to gravity

Divided equation (II) by equation (I)

$\dfrac{g'}{g}=\dfrac{\dfrac{GmM}{(R-\dfrac{R}{2})^2}}{\dfrac{GmM}{(R+\dfrac{R}{2})^2}}$

$\dfrac{g'}{g}=\dfrac{(\dfrac{R}{2})^2}{(\dfrac{3R}{2})^2}$

$\dfrac{g'}{g}=\dfrac{1}{9}$

Hence, The ratio of the acceleration due to gravity is 1:9.

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Topic : acceleration due to gravity

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