Circular Motion and Universal Law of Gravitation Solutions Check Requested

Using the given values for Msun and Rmoon, we can calculate the acceleration on the surface of a white dwarf star:a = (6.67426x10^-11 m^3 kg^-1 s^-2)(1.99x10^30 kg)/(1.74x10^6 m)^2= 4.38x10^7 m/s^2.In summary, the surface acceleration due to gravity on a white dwarf star, with a mass equal to that of our Sun and a radius equal to that of our Moon, is 4.38x10^7 m/s^2. This is significantly higher than the surface acceleration on Earth, which is 9.8 m/s^2
  • #1
rain_ex
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Homework Statement



1. A typical white dwarf star, which once was an average star like our Sun but now in the last stage of its evolution, is the size of our Moon but has the mass of our Sun. What is the surface acceleration due to gravity on this star? (Msun = 1.99 x 10^30 kg, Rmoon = 1.74 x 10^6 m)

- R = radius and M = mass (in reference to Msun and Rmoon)

Homework Equations



Universal Law of Gravitation
0f36df929ac9d711a8ba8c5658c3bfee.png


The Attempt at a Solution



I don't know how to start, but still gave it a try.
First of all, I didn't know what the "F" on the other side of the equation is supposed to be or set the equation up to find the acceleration.
I started by substituting G for 6.67426x10^-11 and got rid of the 2nd mass altogether.
I plugged in the mass of the sun, 1.99x10^30 kg divided by the radius of the moon, 1.74x10^6 m.
Thus I ended up with 4.38x10^7 m/s^2.

Is this the solution?
 
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  • #2
Your try was a good one, well figured out.

We have [tex]F = m_2 a = G \frac{m_1 m_2}{r^2},[/tex]
so [tex]a = G \frac{m_1}{r^2}.[/tex]
 
1.

What is circular motion and how does it relate to the universal law of gravitation?

Circular motion is the movement of an object along a circular path. The universal law of gravitation states that every object in the universe attracts every other object with a force that is directly proportional to their masses and inversely proportional to the square of the distance between them. In circular motion, this force of attraction, known as gravity, acts as the centripetal force that keeps the object moving in a circular path.

2.

How do you calculate the centripetal force in circular motion?

The centripetal force can be calculated using the formula Fc = mv^2/r, where Fc is the centripetal force, m is the mass of the object, v is the velocity of the object, and r is the radius of the circular path. This formula is derived from the universal law of gravitation.

3.

What is the role of gravity in circular motion?

Gravity plays a crucial role in circular motion as it provides the centripetal force that keeps the object moving in a circular path. Without gravity, the object would travel in a straight line instead of a circular path.

4.

How does the radius of the circular path affect the centripetal force?

The centripetal force is inversely proportional to the radius of the circular path. This means that as the radius increases, the centripetal force decreases. This is why objects in larger circular orbits, such as planets around the sun, have a lower centripetal force compared to objects in smaller circular orbits.

5.

Can the universal law of gravitation be applied to objects on Earth's surface?

Yes, the universal law of gravitation can be applied to objects on Earth's surface. However, for objects on Earth's surface, the force of gravity is often simplified to the formula Fg = mg, where Fg is the force of gravity, m is the mass of the object, and g is the acceleration due to gravity. This is because the distance between objects on Earth's surface is relatively small compared to the distance between objects in space, making the inverse square relationship less significant.

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