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Class 9
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Physics
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Gravitation
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Mass of Earth

Gravitation

Mass of Earth

Physics

STATEMENT:

Consider a body of mass mmm is placed at the surface of the Earth. Let,

MEM_EME​ = mass of earth.

RER_ERE​ = radius of earth, Which also is distance between the center of earth and the body.

GGG = Universal gravitational constant.

DERIVATION:

By the Newton’s law of universal gravitation, the force between the body and the earth is given by:

F=GmMERE2\boxed{F =G\frac{mM_E}{{R_E}^2}}F=GRE​2mME​​​ ……….. (i)

But the force with which the earth attracts a body towards its center is called weight of the body, i.e.

F=W=mgF = W=mgF=W=mg …………….. (ii)

Comparing equation (i) and (ii):

mg=GmMERE2mg =G\frac{mM_E}{{R_E}^2}mg=GRE​2mME​​

g=GMERE2g =G\frac{M_E}{{R_E}^2}g=GRE​2ME​​

By re-arranging we get:

ME=gRE2G\boxed{M_E =g\frac{{R_E}^2}{G}}ME​=gGRE​2​​

Here,

g=9.8 m/s2g = 9.8 \thinspace m/s²g=9.8m/s2 G=6.67 x 10−11 Nm2/kg2G = 6.67 \thinspace \text{x} \thinspace 10^{-11} \thinspace Nm^2/kg^2G=6.67x10−11Nm2/kg2

RE=6.38 x 106 mR_E = 6.38 \thinspace \text{x} \thinspace10^6\thinspace mRE​=6.38x106m

By substituting these values, we get:

ME=(9.8)(6.38 x 106)26.67 x 10−11M_E =(9.8) \Large \frac{(6.38 \thinspace \text{x} \thinspace10^6)^2}{6.67 \thinspace \text{x} \thinspace 10^{-11}}ME​=(9.8)6.67x10−11(6.38x106)2​

ME=6.0 x 1024 kg\boxed{M_E = 6.0 \thinspace \text{x} \thinspace 10^{24} \thinspace kg}ME​=6.0x1024kg​