IIT JEE , AIEEE Forum - Physics , Mechanics

aggmegha90

a spherical hole of radius R/2 is excavated from the asteroid of mass M as shown. the gravitational accleration at a point on surface of asteroid just above the excavation is

aggmegha90

any one??

harsha_27

It is $\frac{GM}{2R^2}$ .

aggmegha90

right. tell me how? i will rate u!

harsha_27

Gravity at that point = Gravity due to the complete Asteroid(spherical) at that point - Gravity due to the removed part at that point.

$so \ g_{net} = \frac{GM}{R^2} - \frac{G(\frac{M}{8})}{(\frac{R}{2})^2} \ (Mass \ of \ removed \ part \ = \frac{M}{8})\\ \\<br/>=> g_{net} = \frac{GM}{2R^2}$

aggmegha90

why do we substract the gravity due to complete Asteroid at that point and Gravity due to the removed part
what if we have to find acc due to gravity at any other point??

sagarvaze

why to do so much use gausses law for gravity

integral(g.dA)=4piGm

sagarvaze

so we get

g.4pir^2=4piGm

so gravity at any point = GM/2r^2

harsha_27

Gravity arises due to mass(that is matter).

When we are removing some mass,obviously we have to subtract.

And regarding Gravity at any other point,we have to do the above mentioned subtraction vectorially.

The question is framed at that point to make the calculation easy.

Note:Please avoid phrases like 'If u can" , 'if u have guts'.... .That will boil the blood.

aggmegha90

THANKS!

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a spherical hole of radius R/2 is excavated from the asteroid of mass M as shown. the gravitational accleration at a point on surface of asteroid just above the excavation is