Question

find the ratio of acceleration due to gravity at some height and at depth​

Answer 1

Answer:

Acceleration due to gravity height h,

\displaystyle g_1= g \left( 1-\frac{2h}{R} \right)g

1

=g(1−

R

2h

)

Acceleration due to gravity at depth h,

\displaystyle g_2= g \left( 1-\frac{h}{R} \right)g

2

=g(1−

R

h

)

\therefore∴ \displaystyle \frac{g_1}{g_2} =\frac{1-2h/R}{1-h/R}

g

2

g

1

=

1−h/R

1−2h/R

\displaystyle = \left( 1-\frac{2h}{R} \right) \left( 1-\frac{h}{R} \right)^{-1}=(1−

R

2h

)(1−

R

h

)

−1

\displaystyle =\left( 1-\frac{h}{R} \right)=(1−

R

h

)

[neglecting higher power of \displaystyle \frac{h}{R}

R

h

]

\displaystyle \therefore \frac{g_1}{g_2}∴

g

2

g

1