It can be shown that for a uniform sphere the force of gravity at a point inside the sphere depends only on the mass closer to the center than that point. The net force of gravity due to points outside the radius of the point cancels.
Part A
How far would you have to drill into the Earth, to reach a point where your weight is reduced by 8.0\% ? Approximate the Earth as a uniform sphere.
g= GMe/re^2
g'=.92GMe/re^2 =GMe(r/re)^3
then it reduces to
.92re=r^3, and that leads to
r=re *cubrt(.92)
the r is, 6205.11, and shouldn' t I need to subratc it from radius of earth which is 6380-6205.11, it is equal to "174.89" but my answer fails. Please help, ty for ur care now.
To find the distance you would have to drill into the Earth to reach a point where your weight is reduced by 8%, you correctly started with the equation:
g' = g * (r / Re)^3
Where g is the acceleration due to gravity at the surface and Re is the radius of the Earth. You approximated the Earth as a uniform sphere, denoted by re.
You then substituted g' with 0.92g, since your weight would be reduced by 8%, and simplified the equation to:
0.92g = g * (r / Re)^3
Next, you rearranged the equation to solve for r:
0.92 = (r / Re)^3
Taking the cube root of both sides gives:
cubert(0.92) = r / Re
Now, to find the value of r, you multiplied Re by the cube root of 0.92:
r = Re * cubert(0.92)
Using the radius of the Earth, Re = 6380 km, and evaluating the expression, you correctly obtained r ≈ 6205.11 km.
However, to find the distance you would have to drill into the Earth, you need to subtract this result from the radius of the Earth, not the other way around:
Distance drilled = Re - r = 6380 km - 6205.11 km ≈ 174.89 km
So, the correct distance you would have to drill into the Earth is approximately 174.89 km.