 # Acceleration due to Gravity

• The variation of acceleration due to gravity with height can be calculated at \tt gh=\frac{GM}{\left(R+h\right)^2}
• Fractional decrease in ‘g’ for smaller attitude is \tt \frac{\Delta g}{g}=\frac{2h}{R} which is used only for very small value.
• Acceleration due to gravity below the surface of earth can be calculated by gd = g(1- d/R) where ‘d’ is the depth from surface.
• At the centre of the earth acceleration due to gravity is ‘0’
• Acceleration due to gravity at a height and at a depth are equal when 2h = d.
• Variation of acceleration due to gravity when the earth rotates with an angular velocity ‘w’ and a latitude angle φ is gφ = g - Rw2 cos2φ.
• The value of ‘g’ the equator depends on angular velocity of earth.
• The value of ‘g’ at poles is independent of the rotation of the earth.
• The value of ‘g’ at equator becomes zero when the angular velocity of rotation increases to 17 times the present value.
• The value of ‘g’ gradually increases from equator to poles.
• The value of ‘g’ is slightly more at the location of minerals deposits.
• The value of ‘g’ is slightly less on the top of mountain and also inside mines.
• If g1 g2 g3 are the acceleration due to gravity's on the surface of earth on the top of a mountain and inside a mine then g2 > g1 > g3.

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1. Consider a body of mass m, lying on the surface of earth then gravitational force on the body is given by \tt F=\frac{GMm}{R^{2}}

2. Acceleration due to gravity \tt g=\frac{GM}{R^{2}}

3. Acceleration due to gravity at height h from the surface of the earth \tt g'=\frac{GM}{\left(R+h\right)^2}

4. As we go above the surface of the earth, the value of g decreases because \tt g'\propto\frac{1}{r^{2}}

5. Acceleration due to gravity at depth d from the surface of the earth g'=\frac{4}{3}\pi \rho G\left(R-d\right)