The mass of Earth is 5.97 ×10^24 kg and its orbital radius is an average of 1.50 ×10^11 m. Calcu

The mass of Earth is 5.97 ×10^24 kg and its orbital radius...

The mass of Earth is $5.97 \times 10^{24} \mathrm{kg}$ and its orbital radius is an average of $1.50 \times 10^{11} \mathrm{m}$. Calculate the magnitude of its linear momentum at the location in the diagram.

Key Concepts

Linear Momentum

Linear momentum is a fundamental concept in mechanics, defined as the product of an object's mass and its velocity. It is a vector quantity that represents the amount of motion an object possesses, and it is conserved in isolated systems, making it crucial for analyzing collisions and orbital dynamics.

Orbital Motion

Orbital motion refers to the movement of a body in a curved trajectory around a central object, typically due to gravitational attraction. This concept involves understanding how the balance between an object's inertia and the gravitational pull of the central body results in a stable path, often approximated as circular or elliptical.

Gravitational Force and Centripetal Acceleration

Gravitational force, as described by Newton's Law of Universal Gravitation, provides the necessary centripetal force that keeps a body in orbit. The centripetal acceleration required for circular motion is achieved when this gravitational force perfectly balances the inertia of the moving object, thereby determining its orbital velocity and related momentum.

Transcript

00:01 So we're calculating the magnitude of the linear momentum of the earth where it's at the position in this is some diagram that i can't see.

00:08 So i'm going to imagine that we've got the sun here, right, all 1 .99 times 10 to the 30th kilograms of the sun, right? and we've got the earth going in its orbit, right? so let's imagine the earth is there.

00:25 It's got a velocity of that direction, right? and here we've got 5 .9, 7 times 10 to the 24th kilograms, right? the momentum is going to be mv, but the question is, you know, what's the velocity of the earth as it goes around the sun? to solve that, we would, we would, let's assume that the sun doesn't move, right? it's a million times, almost a million times more massive than the earth, right? so what we can do is we can do a little maths here, right? we know that the force of gravity, mass of the sun, mass of the earth over r squared is going to be the mass of the earth times v squared over r.

01:05 Right.

01:06 So this is centripetal force equals gravitational force.

01:09 The earth mass cancels.

01:11 We can cancel a radius here.

01:13 And we end up with the velocity being a square root of g times the mass of the sun over r, right? we can plug some numbers into that.

01:23 Right so our velocity of the earth's and we'll figure out the velocity of the earth's orbit here right it's going to be the square root of 6 .67 times 10 to the minus 11th right and then the key here is we've got to use the sun's mass right because the earth's mass canceled right so 1 .99 times 10 to the 30th right and then they said 1 .5 times 10 to the 11th meters is our are distanced from the sun, right? let's figure that out.

01:59 It's going to take some math here.

02:02 Square root of 6 .67 times 10 to the minus 11th times 1 .99 times 10 to the 30th, divided by 1 .5 times 10 to the 11th...

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