ToughSTEM
ToughSTEM
A question answer community on a mission
to share Solutions for all STEM major Problems.
Cant find a problem on ToughSTEM?
0

Suppose you drill a tunnel through the earth along a diameter and drop a stone of mass m=13.47 kg down the tunnel. The force of gravity within the volume of Earth is radially inward and equals to:

$F_{g}= -\frac{GMm}{R^{3}}r$

where r is the radial distance from the center of earth. R, G, M are constants with their values shown below. Assuming the density of Earth is constant and no friction or air resistance in the tunnel, the stone will travel all the way until it reaches the other end of the tunnel then travel back to your hand under the effect of the restoring gravitational force in the above equation.

How long (in minutes) is the journey of the stone from the moment you drop it off to the moment when you catch it?

The mass of Earth is M=5.97258*10^24 kg, Radius of Earth is R=6.306*10^6 m, and the gravitational constant is G=6.67384*10^-11 m^3/kg s^2.

Edit
Community
1
Comment
Solutions
0

initial veocity=0

$F_{g}= -\frac{GMm}{R^{3}}r$= -k*r where k=GMm/R3

so accelartion=-GMr/R3

now as we know for an SHM accelaration is directly propotional to distance from equilibrium with a negative sign.

and for SHM a time period will be T=$2\pi\sqrt(\frac{m}{k})$=$2\pi\sqrt(\frac{R^{3}}{GM})$

=2*3.14*((6.306*10^6)3/(6.67384*10^-11*5.97258*10^24))0.5=4981sec=83min 1sec

Edit
Community
1
Comment
Close

Choose An Image
or
Get image from URL
GO
Close
Back
Close
What URL would you like to link?
GO
α
β
γ
δ
ϵ
ε
η
ϑ
λ
μ
π
ρ
σ
τ
φ
ψ
ω
Γ
Δ
Θ
Λ
Π
Σ
Φ
Ω
Copied to Clipboard

to interact with the community. (That's part of how we ensure only good content gets on ToughSTEM)
OR
OR
ToughSTEM is completely free, and its staying that way. Students pay way too much already.
Almost done!