Will a object falling from a height $x$ above the surface experience the same impact as another object falling In a hole of depth $x$ from the surface?
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$\begingroup$ You need to find out how the gravitational field strength varies with position above and below the Earth’s surface. $\endgroup$– FarcherCommented Jan 21, 2018 at 6:29
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$\begingroup$ If x is not comparable to R(radius of earth) then yes. There will be a negligible change in g. $\endgroup$– HyperBeanCommented Jan 21, 2018 at 6:29
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$\begingroup$ @HyperBean will the impact on both the objects be the same after hitting their respective grounds? $\endgroup$– user160598Commented Jan 21, 2018 at 6:49
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$\begingroup$ That depends mainly on what you define to be impact. If its the force acting on the body just during contact or the change in momenta just before and just after then it depends upon the coefficient of restitution. $\endgroup$– HyperBeanCommented Jan 21, 2018 at 6:51
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$\begingroup$ @HyperBean I mean force acting on the body during the contact $\endgroup$– user160598Commented Jan 21, 2018 at 6:52
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3 Answers
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No. Both will experience different forces. By the law of gravitation, $$ F = G\frac{Mm}{R^2} $$ So since the difference from the center of mass of the earth is different in both cases the forces are different. This means both will have different velocities when they have an impact and therefore the impact force will be different.
This difference is usually ignored if the distance x is very small compared to R because the difference in forces in both the cases will be negligible.
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1$\begingroup$ The force experienced by an object in a hole is not $\frac{GMm}{r^2}$, but $\frac{GMmr}{R^3}$, where $R$ is the radius of the earth and $r$ is the distance from the center of the earth. $\endgroup$– Chris ♦Commented Jan 21, 2018 at 8:13
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The value of g will change slightly given that the distance from the Earth's center will increase (if the object falls from height X above the ground) or decrease (if the object falls from the ground into a hole of depth x). This is due to the inverse square law given for finding the value of the force of gravity exerted by Earth onto the object. However, unless x is significantly large in comparison to Earth's radius, the change in g will be insignificant and the value 9.8 m/s/s will work in an applied calculation.
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If X id not big enough.It won't make a difference. Change in the value of g will be too small.
answered Jan 21, 2018 at 6:58