Menu Close

Imagine-that-a-hole-is-drilled-from-one-point-on-the-earth-surface-to-the-other-side-through-the-diameter-of-the-earth-and-small-of-mass-m-released-into-the-hole-show-that-the-hole-of-mass-exc




Question Number 177240 by peter frank last updated on 02/Oct/22
Imagine that a hole is drilled  from one point on the earth   surface to the other side through the  diameter of the earth and small  of mass ^′ m^′  released into the hole.  show that the hole of mass excute  simple harmonic motion and  find its period of revolution
$$\mathrm{Imagine}\:\mathrm{that}\:\mathrm{a}\:\mathrm{hole}\:\mathrm{is}\:\mathrm{drilled} \\ $$$$\mathrm{from}\:\mathrm{one}\:\mathrm{point}\:\mathrm{on}\:\mathrm{the}\:\mathrm{earth}\: \\ $$$$\mathrm{surface}\:\mathrm{to}\:\mathrm{the}\:\mathrm{other}\:\mathrm{side}\:\mathrm{through}\:\mathrm{the} \\ $$$$\mathrm{diameter}\:\mathrm{of}\:\mathrm{the}\:\mathrm{earth}\:\mathrm{and}\:\mathrm{small} \\ $$$$\mathrm{of}\:\mathrm{mass}\:\:^{'} \mathrm{m}^{'} \:\mathrm{released}\:\mathrm{into}\:\mathrm{the}\:\mathrm{hole}. \\ $$$$\mathrm{show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{hole}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{excute} \\ $$$$\mathrm{simple}\:\mathrm{harmonic}\:\mathrm{motion}\:\mathrm{and} \\ $$$$\mathrm{find}\:\mathrm{its}\:\mathrm{period}\:\mathrm{of}\:\mathrm{revolution} \\ $$
Commented by JDamian last updated on 02/Oct/22
What is "the hole of mass"?
Commented by mr W last updated on 02/Oct/22
he means the “mass in the hole”,   i think.
$${he}\:{means}\:{the}\:“{mass}\:{in}\:{the}\:{hole}'',\: \\ $$$${i}\:{think}. \\ $$
Answered by mr W last updated on 02/Oct/22
Commented by mr W last updated on 02/Oct/22
assume that the earth is an uniform  solid sphere with density ρ and   radius R.  when the small mass m in the hole  is at a distance x to the center of  the earth, it is attracted by the  gravitational forceF of the earth.  the gravitational force from the part  of the earth inside the radius x   (mass M_1 ) is  F_1 =((GmM_1 )/x^2 )  with M_1 =((4πx^3 ρ)/3)  ⇒F_1 =((Gm)/x^2 )×((4πx^3 ρ)/3)=((4πGmρx)/3)  the gravitational force from the part  of the earth outside the radius x  (mass M_2 ) is F_2 =0.  F=F_1 +F_2 =((4πGmρx)/3)  the acceleration of the small mass is  a=(d^2 x/dt^2 ).  we have  ma=−F=−((4πGmρx)/3)  with k=((4πGρ)/3)=constant  ⇒(d^2 x/dt^2 )+kx=0  this is the equation of a simple  harmonic motion.  ω=(√k)=(√((4πGρ)/3)).  g=((GM)/R^2 )=((4πGρR)/3)=kR  ⇒k=(g/R)  ⇒ω=(√(g/R))  the period is T=((2π)/ω)=2π(√(R/g))  with R=6381 km, g=9.81 m/s^2   we get   T=2π(√((6381000)/(9.81)))=5067 s ≈1h24m
$${assume}\:{that}\:{the}\:{earth}\:{is}\:{an}\:{uniform} \\ $$$${solid}\:{sphere}\:{with}\:{density}\:\rho\:{and}\: \\ $$$${radius}\:{R}. \\ $$$${when}\:{the}\:{small}\:{mass}\:{m}\:{in}\:{the}\:{hole} \\ $$$${is}\:{at}\:{a}\:{distance}\:{x}\:{to}\:{the}\:{center}\:{of} \\ $$$${the}\:{earth},\:{it}\:{is}\:{attracted}\:{by}\:{the} \\ $$$${gravitational}\:{forceF}\:{of}\:{the}\:{earth}. \\ $$$${the}\:{gravitational}\:{force}\:{from}\:{the}\:{part} \\ $$$${of}\:{the}\:{earth}\:{inside}\:{the}\:{radius}\:{x}\: \\ $$$$\left({mass}\:{M}_{\mathrm{1}} \right)\:{is} \\ $$$${F}_{\mathrm{1}} =\frac{{GmM}_{\mathrm{1}} }{{x}^{\mathrm{2}} } \\ $$$${with}\:{M}_{\mathrm{1}} =\frac{\mathrm{4}\pi{x}^{\mathrm{3}} \rho}{\mathrm{3}} \\ $$$$\Rightarrow{F}_{\mathrm{1}} =\frac{{Gm}}{{x}^{\mathrm{2}} }×\frac{\mathrm{4}\pi{x}^{\mathrm{3}} \rho}{\mathrm{3}}=\frac{\mathrm{4}\pi{Gm}\rho{x}}{\mathrm{3}} \\ $$$${the}\:{gravitational}\:{force}\:{from}\:{the}\:{part} \\ $$$${of}\:{the}\:{earth}\:{outside}\:{the}\:{radius}\:{x} \\ $$$$\left({mass}\:{M}_{\mathrm{2}} \right)\:{is}\:{F}_{\mathrm{2}} =\mathrm{0}. \\ $$$${F}={F}_{\mathrm{1}} +{F}_{\mathrm{2}} =\frac{\mathrm{4}\pi{Gm}\rho{x}}{\mathrm{3}} \\ $$$${the}\:{acceleration}\:{of}\:{the}\:{small}\:{mass}\:{is} \\ $$$${a}=\frac{{d}^{\mathrm{2}} {x}}{{dt}^{\mathrm{2}} }. \\ $$$${we}\:{have} \\ $$$${ma}=−{F}=−\frac{\mathrm{4}\pi{Gm}\rho{x}}{\mathrm{3}} \\ $$$${with}\:{k}=\frac{\mathrm{4}\pi{G}\rho}{\mathrm{3}}={constant} \\ $$$$\Rightarrow\frac{{d}^{\mathrm{2}} {x}}{{dt}^{\mathrm{2}} }+{kx}=\mathrm{0} \\ $$$${this}\:{is}\:{the}\:{equation}\:{of}\:{a}\:{simple} \\ $$$${harmonic}\:{motion}. \\ $$$$\omega=\sqrt{{k}}=\sqrt{\frac{\mathrm{4}\pi{G}\rho}{\mathrm{3}}}. \\ $$$${g}=\frac{{GM}}{{R}^{\mathrm{2}} }=\frac{\mathrm{4}\pi{G}\rho{R}}{\mathrm{3}}={kR} \\ $$$$\Rightarrow{k}=\frac{{g}}{{R}} \\ $$$$\Rightarrow\omega=\sqrt{\frac{{g}}{{R}}} \\ $$$${the}\:{period}\:{is}\:{T}=\frac{\mathrm{2}\pi}{\omega}=\mathrm{2}\pi\sqrt{\frac{{R}}{{g}}} \\ $$$${with}\:{R}=\mathrm{6381}\:{km},\:{g}=\mathrm{9}.\mathrm{81}\:{m}/{s}^{\mathrm{2}} \\ $$$${we}\:{get}\: \\ $$$${T}=\mathrm{2}\pi\sqrt{\frac{\mathrm{6381000}}{\mathrm{9}.\mathrm{81}}}=\mathrm{5067}\:{s}\:\approx\mathrm{1}{h}\mathrm{24}{m} \\ $$
Commented by peter frank last updated on 02/Oct/22
thank you.I understand  very well
$$\mathrm{thank}\:\mathrm{you}.\mathrm{I}\:\mathrm{understand}\:\:\mathrm{very}\:\mathrm{well} \\ $$
Commented by mr W last updated on 03/Oct/22
it means when you could take this  short cut, you would be able to travel  from the north pole to the south pole  in only 42 minutes!
$${it}\:{means}\:{when}\:{you}\:{could}\:{take}\:{this} \\ $$$${short}\:{cut},\:{you}\:{would}\:{be}\:{able}\:{to}\:{travel} \\ $$$${from}\:{the}\:{north}\:{pole}\:{to}\:{the}\:{south}\:{pole} \\ $$$${in}\:{only}\:\mathrm{42}\:{minutes}! \\ $$
Commented by peter frank last updated on 03/Oct/22
thank you
$$\mathrm{thank}\:\mathrm{you} \\ $$
Commented by Tawa11 last updated on 03/Oct/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *