Theres something about gravitational

[ACLerok]

With only two lectures in my back pocket, I still can't find a reasonable solution for this problem.

Suppose that a planet were discovered between the sun and Mercury, with a circular orbit of radius equal to 2/3 of the average orbit radius of Mercury. The orbital period of Mercury is 88.0 days.

What would be the orbital period of such a planet?

and also, in general:

Four identical masses of mass 500 kg each are placed at the corners of a square whose side lengths are 15. cm.

What is the magnitude of the net gravitational force on one of the masses, due to the other three?

 

[jamesrc]

First one:
You can use Kepler's Third Law (or equate the centripetal force with the gravitational attractio if you'd rather):

\frac{T_1^2}{R_1^3} = \frac{T_2^2}{R_2^3}

With T₁ = 88 days and R₂ = 2*R₁/3 to find the period of the mysterious new planet.

Second one:

Just do a little vector addition here. You know that the gravitational attraction between two masses separated by a distance r is given by

\vec{F_{12}} = -\frac{{\rm G}m_1 m_2}{{\vec{|r_{12}|}}^3}\vec{r_{12}}

Just so we have our notation straight, F₁₂ indicates the force on 1 due to 2, while r₁₂ indicates the position vector from 1 to 2.

Just sum the forces (good old law of superposition). Take the top left mass as an example: you'll have a force pointing down due to the bottom left mass, a force pointing to the right due to the top right mass, and a force pointing diagonally toward the bottom right mass due to that mass. The first two forces will have the same magnitude, while the third force will have a magnitude of half of either of the first two forces.

Hint: the net force on each mass in this configuration should be directed radially inward to the center of the square the 4 masses make.

 

[M98Ranger]

Gravitational force is a very complex and fascinating phenomenon that has been studied by scientists for centuries. It is understandable that with only two lectures under your belt, finding a reasonable solution to a problem involving gravitational force can be challenging. However, with more practice and understanding, you will be able to tackle such problems confidently.

To answer your first question, the orbital period of the planet between the sun and Mercury can be calculated using Kepler's third law, which states that the square of the orbital period is directly proportional to the cube of the average distance between the planet and the sun. Since the planet has a circular orbit with a radius 2/3 of Mercury's, the average distance would be 2/3 of the distance between the sun and Mercury, which is 0.5 AU. Plugging in the values, we get an orbital period of approximately 49.3 days.

For the second question, the magnitude of the net gravitational force on one of the masses can be calculated using Newton's law of universal gravitation, which states that the force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. In this case, the distance between the masses is 15 cm, and the mass of each object is 500 kg. Plugging in the values, we get a net gravitational force of approximately 1.33 x 10^-6 N.

In general, solving problems involving gravitational force requires a good understanding of the fundamental principles and equations. It also requires practice and patience. As you continue to learn and explore this fascinating topic, you will be able to find reasonable solutions to more complex problems. Keep up the good work!