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Suppose a large spherical object, such as a planet, with radius $R$ and mass $M$ has a narrow tunnel passing diametrically through it. A particle of mass $m$ is inside the tunnel at a distance $x≤R$ from the center. It can be shown that the net gravitational force on the particle is due entirely to the sphere of mass with radius $r≤x$; there is no net gravitational force from the mass in the spherical shell with $r>x$.

$\left(a\right)$ Find an expression for the gravitational force on the particle, assuming the object has uniform density. Your expression will be in terms of $x,R,m,M$, and any necessary constants.

$\left(b\right)$ You should have found that the gravitational force is a linear restoring force. Consequently, in the absence of air resistance, objects in the tunnel will oscillate with $SHM$. Suppose an intrepid astronaut exploring a $150-km-$diameter, $3.5×{10}^{18}kg$ asteroid discovers a tunnel through the center. If she jumps into the hole, how long will it take her to fall all the way through the asteroid and emerge on the other side?

Step by step solution

01

Step: 1 Gravitational force:

Newton's law of gravity can be used to describe the gravitational force between two bodies. The force felt by anyone under the gravity of another body is equal to the ratio of their masses and directly proportional to the distance of their distance. The gravitational force is expressed as:

$F=G\frac{mM}{x^2}$

02

Step; 2 Expression for gravitational force: (part a)

The gravitational force as

$F=G\frac{m{M}_{in}}{x^2}$

The total mass of sphere is

$M_{in}=M\frac{x^3}{R^3}$

The partial mass force by

$F=G\frac{mM}{R^3}x$

This shows restoring Hooke's law and linear in force.

03

Step: 3 Finding gravitational: (part b)

The duration of the vibrating is the time it takes the body to complete one full oscillation. The timeframe of vibration is expressed as:

$T=2\mathrm{Ï€}\sqrt{\frac{m}{k}}$

The astronaut emerges on the opposite side in half the time it takes to travel to another side.

The time period expression by

$\begin{array}{l}{T}_{\text{half}}=2\mathrm{Ï€}\sqrt{\frac{m}{k}}\\ T_{\text{half}}=\mathrm{Ï€}\sqrt{\frac{m}{GmM/R^3}}\\ T_{\text{half}}=\mathrm{Ï€}\sqrt{\frac{R^3}{GM}}\end{array}$

04

Step; 4 Substituting: (part b)

Substituting and getting values for $75km$inabove equation as,

$\begin{array}{l}{T}_{\text{half}}=\mathrm{π}\sqrt{\frac{(75\mathrm{km}{)}^3}{\left(6.67×{10}^{−11}{\mathrm{N}}^2{\mathrm{mg}}^2\right)\left(3.5×{10}^{18}\mathrm{kg}\right)}}\\ T_{\text{half}}=4200\mathrm{s}\\ T_{\text{half}}=70\min .\end{array}$

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