Calculate Force and Period of Asteroid Orbit Around Sun | Physics Help ASAP!

[o04jjang]

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Four 1,000,000- kg masses are arranged in a square, 74meters on a side. To the nearest ten thousandth of a Newton, what is the magnitude of the force the mass in the lower left hand corner?

If the Earth's period is 365.25 days and its distance to the (center of the) Sun is one astronomical unit (AU), what is the period, to the nearest day, of an asteroid in a circular orbit about the Sun with a radius of AU?

 

[learningphysics]

Four 1,000,000- kg masses are arranged in a square, 74meters on a side. To the nearest ten thousandth of a Newton, what is the magnitude of the force the mass in the lower left hand corner?

You've got 3 vectors... The force due to the top left hand mass on the lower left hand mass. The force due to the top right hand mass on the lower left hand mass. And the force due to the left right hand mass on the lower left hand mass.


I'd do it this way. Get the total vertical component of the force on the lower left hand mass first. That means the force due to the top left hand mass (which is vertically downward) with the VERTICAL component of the top right hand mass (so you'll need a little trigonometry here).

Then I'd get the total horizontal component. That means the force due to the bottom right hand mass (which is horizontal leftward) along with the HORIZONTAL component of the top right hand mass.

The two components are the same value... by symmetry. So you can actually just get the vertical, and then you know the horizontal is the same thing.

When you've gotten the two components, use the pythagorean theorem to get the total magnitude.

If the Earth's period is 365.25 days and its distance to the (center of the) Sun is one astronomical unit (AU), what is the period, to the nearest day, of an asteroid in a circular orbit about the Sun with a radius of AU?

Is the radius of the asteroid's orbit also 1 AU? You just wrote radius of AU but I want to make sure you didn't mistype.

 

[thepatientmental]

To calculate the force on the mass in the lower left hand corner, we can use the formula for gravitational force: F = G(m1m2)/r^2, where G is the gravitational constant (6.67x10^-11 Nm^2/kg^2), m1 and m2 are the masses of the two objects, and r is the distance between them.

In this case, m1 = 1,000,000 kg, m2 = 1,000,000 kg, and r = 74 meters. Plugging these values into the formula, we get:

F = (6.67x10^-11 Nm^2/kg^2)(1,000,000 kg)(1,000,000 kg)/(74 meters)^2
= 0.0116 N

Therefore, the magnitude of the force on the mass in the lower left hand corner is approximately 0.0116 Newtons.

To calculate the period of an asteroid in a circular orbit around the Sun, we can use Kepler's third law: T^2 = (4π^2r^3)/(GM), where T is the period, r is the radius of the orbit, G is the gravitational constant, and M is the mass of the Sun.

In this case, r = 1 AU and M = 1.989x10^30 kg (mass of the Sun). Plugging these values into the formula, we get:

T^2 = (4π^2)(1 AU)^3/(6.67x10^-11 Nm^2/kg^2)(1.989x10^30 kg)
= 2.97x10^20 seconds^2

Taking the square root of both sides, we get:

T = √(2.97x10^20 seconds^2)
= 1.72x10^10 seconds

Converting this to days, we get:

T ≈ 1.72x10^10 seconds x (1 day/86,400 seconds)
≈ 199,074 days

Therefore, the period of an asteroid in a circular orbit around the Sun with a radius of 1 AU is approximately 199,074 days to the nearest day.