Tamil Nadu Board of Secondary EducationHSC Science Class 11

# Acceleration Due to Gravity of the Earth

#### Topics

• Variation of g with altitude, depth and latitude

## Notes

#### Acceleration due to gravity of the earth

• Acceleration attained due to gravity of earth.

• All the objects fall towards the earth because of gravitational pull of the earth.

• And when abody is falling freely, it will have some velocity and therefore it will attain some acceleration. This acceleration is known as acceleration due to gravity.

• It is a vector quantity.

• Denoted by ‘g’.

• Its value is 9.8  m//s^2.

Example:-Stones falling from a rock will have some velocity because of which some acceleration. This acceleration is due to the force exerted by the earth on the rocks.This is known as acceleration due to gravity.

#### Expression for Acceleration due to gravity

• Consider any object of mass ‘m’ at a point A on the surface of the earth.

• The force of gravity between the body and earth can be calculated as

"F" ="G m M_e/R_e^2       ...(1)

m=mass of the body

Me=mass of the Earth

R= distance between the body and the earth is same as the radius of the earth

F = ma              ....(2)

from eq (1) & (2)

"g"="F"/"m"="GM"_e/"R"_e^2

(1)Problem: Calculate the acceleration due to gravity on the (a) earth's surface (b) at a height of 1.5 xx10^5m from the earths surface.

Solution:

Given- "R"_e = 6.4xx10^6m; "M"_e=5.98xx10^24kg

(a) "g"="GM"_e/"R"_e^2

=(6.67xx10^-11 xx5.98xx10^24)/(6.4xx10^6)^2

=9.74m//"s"^2

(b)Given:  "h"=1.5xx10^5m

"g"("h")="GM"_e/(R_e+h)^2

=(6.67xx10^-11 xx5.98xx10^24)/((6.4xx10^6+1.5xx10^5))

=9.30  m//s^2

(2)Problem: Calculate the mass of the moon if the free fall acceleration near its surface is known to be 1.62m/s2. Radius of the moon is 1738km?

Solution:

g=1.62m/s2

Rm=1738km=1738 x 103m

"G" = 6.67 xx 10^-11 "N"m^2//kg^2

"g"="GM"_m/"R"_m^2

"M"_m = "gR"_m^2/G

=(1.62xx(1738xx10^3)^2)/(6.67xx10^-11)

=7.34xx10^22kg

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