Period of falling through asteroid vs orbit

In summary, the conversation discusses the period of oscillation and orbit velocity of a spherical asteroid, calculated using its density and Newton's gravitational constant. The two methods of calculation are compared and the connection between "g" and the Universal Law of Gravitation is explored. The use of the letter π instead of the word "pi" is also mentioned. The potential impact of differentiation of the asteroid's density on these calculations is also brought up.
  • #1
mintsnapple
50
0

Homework Statement


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Homework Equations



See below

The Attempt at a Solution


My book derives the period of oscillation through the tunnel:
T = 2pi/w = 2pi*sqrt(m/k) = 2pi*sqrt(3/(4pi*G*p)) = sqrt(3pi/(G*p))
Where p is the density of the asteroid, and G is the Newton's gravitational constant.

I know that the orbit velocity is found by equating the gravity force to the necessary centripetal force:
mg = mv^2/r
v = sqrt(r*g)
So the period is 2*pi*r/v = 2pi*sqrt(r/g)

I know these two periods are equal. Can anyone help me with putting them and similar terms and proving that they are?
 
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  • #2
What is g on that spherical asteroid? Do you see any connection between "g" and the Universal Law of Gravitation?

ehild
 
Last edited:
  • #3
If you click on "Go advanced" you will find that it is easy to use the letter π instead of the word "pi".
 
  • #4
Extra credit: what if the asteroid is differentiated, that is the density at its core is higher than at its mantle.
 
  • #5


The period of oscillation through the asteroid tunnel and the period of orbit around the asteroid are indeed equal. This can be seen by equating the expressions for their respective periods:

T_tunnel = sqrt(3pi/(G*p))
T_orbit = 2pi*sqrt(r/g)

By substituting the expression for velocity in the orbit period equation, we get:

T_orbit = 2pi*sqrt(r/(g*g)) = 2pi*sqrt(r/(GM/r^2)) = 2pi*sqrt(r^3/GM)

We can then equate this to the expression for T_tunnel:

T_tunnel = sqrt(3pi/(G*p)) = sqrt(3pi/(G*(m/V))) = sqrt(3pi/(G*(m/(4/3*pi*r^3)))) = sqrt(3pi/(G*(3m/(4pi*r^3)))) = sqrt(r^3/(G*m))

Since both expressions for the periods are now in terms of r^3/GM, we can set them equal to each other:

sqrt(r^3/GM) = sqrt(r^3/GM)
This proves that the periods are equal and that the time it takes to fall through the asteroid tunnel is the same as the time it takes to orbit around the asteroid. This result is intuitive, as both situations involve the same gravitational force and the same distance from the center of the asteroid.
 

FAQ: Period of falling through asteroid vs orbit

What is the difference between falling through an asteroid and orbiting it?

The main difference between falling through an asteroid and orbiting it is the path that the object takes. When an object falls through an asteroid, it follows a straight path towards the center of the asteroid. On the other hand, when an object orbits an asteroid, it follows a curved path around the asteroid, constantly being pulled towards its center by gravity.

How does the size of the asteroid affect the period of falling through it vs orbiting it?

The size of the asteroid has a significant impact on the period of falling through it and orbiting it. Smaller asteroids have a lower gravitational pull, which means that objects will fall through them faster and have shorter orbit periods compared to larger asteroids.

What factors determine the speed at which an object falls through an asteroid?

The speed at which an object falls through an asteroid is primarily determined by the mass and size of the asteroid, as well as the initial velocity of the object. The higher the mass and size of the asteroid, the stronger its gravitational pull, and the faster the object will fall. The initial velocity of the object also plays a role, as it can either increase or decrease the speed at which it falls through the asteroid.

Can an object fall through an asteroid and orbit it at the same time?

No, an object cannot fall through an asteroid and orbit it simultaneously. When an object falls through an asteroid, it follows a straight path towards its center, while orbiting an asteroid requires a curved path around its center. These two paths are not compatible and cannot occur at the same time.

How can the period of falling through an asteroid vs orbiting it be calculated?

The period of falling through an asteroid can be calculated using the formula t = √(2d/g), where t is the time taken, d is the distance from the starting point to the center of the asteroid, and g is the gravitational acceleration. The period of orbiting an asteroid can be calculated using the formula T = 2π√(r^3/GM), where T is the period, r is the distance between the object and the center of the asteroid, G is the gravitational constant, and M is the mass of the asteroid.

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