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zubka84 [21]
3 years ago
8

I'm trying to find the mass and radius to help me with this problem: g=(9.81)M/r^2. Is there any way you can help me?

Physics
1 answer:
sp2606 [1]3 years ago
8 0

Answer:

9.81 m/s2

Explanation:

g = 9.81m/s2 means that if you were on Earth, and you somehow were able to create a space free of any air (a vacuum), and you dropped something, then that object’s velocity would be 9.81 m/s after one second, 19.62 after another second, 29.43 after another and so on. In other words, it would accelerate at a rate of 9.81 m/s2.

The reason why it has to be in a vacuum is because the object would be immersed in air otherwise. And the air would cause it to slow down, for the same reason why something dropped in honey takes so long to reach the bottom of the pot: viscosity. The air or the honey would pull on the object as it fell, causing it to experience an acceleration smaller than 9.81 m/s2. In fact, the pull is proportional to the square of the velocity of the falling object, so the faster it gets, the stronger the pull of the air.

At some point, the pull of the air just matches that of gravity, and the object stops accelerating, and falls at a constant velocity for the remainder of its descent. That velocity is called terminal velocity. The effect is much less pronounced in air than in honey because air’s viscosity is much smaller: air flows easily.

It has to be on Earth because the gravitational force changes with distance, and the mass of whatever is pulling, which is Earth. If you were on the moon, g would be different from what it is on Earth, about a sixth of what it is, because of the change in distance, and the change in mass. The actual formula for the force of gravity is called Newton’s law of Universal Gravitation, and is:

Where M and m are the masses of the two objects, G is a universal constant for gravity with a value of 6.67 Nm2/kg2. R is the distance between the two objects, and the minus sign is just there to get the direction right.

g is derived from this formula. M is constant, because it represents the mass of the Earth, and R should change, since it is the distance from the centre of the Earth to the object under consideration, but the Earth is large, so any small change in R due to the object’s height can be neglected, so R can be taken to be a constant, the radius of the Earth. Since g is the acceleration due to gravity on Earth, you can divide F by m to obtain it, since F = ma.

Thus:

Where the minus sign was ignored, because you are more interested in the magnitude of gravity on Earth than its direction. If you wander too far from Earth R will change significantly, so g must change as well, and if you change celestial bodies, M must change.

Thus g = 9.81 m/s2 means that on Earth, in a vacuum, if you were falling, you would accelerate at a rate of 9.81 m/s2, regardless of how your mass.

g can also be taken to mean the strength of the gravitational field. The gravitational field is how the force of gravity gets transferred across space. Gravity does not require the two interacting objects to be in contact with each other, otherwise no life would exist on Earth from touching the sun. There is some distance between Earth and the sun and gravity is transferred through them via the gravitational field.

The gravitational field is defined as the force of gravity per unit mass of a source. The strength of the field is just its magnitude. In other words, the strength is:

Which is exactly the same as the acceleration due to gravity.

The similarity between the strength and the acceleration may not seem to be much, but it is in fact very important. It is for this reason that all falling objects experience the same acceleration.

Gravity is the only force that has this property. If you compared it to the very similar Coulomb force, which is essentially the same as Newton’s gravitational law, but with charges replacing the masses, you see that for different masses, the Coulomb force causes different accelerations, whereas gravity does not. It is this property of gravity that prompted Einstein to formulate the theory of General Relativity, where gravity is described in terms of geometry.

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Using Rayleigh's criterion, calculate the diameter of an earth-based telescope that gives this resolution with 700 nm light.
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Complete Question

Due to blurring caused by atmospheric distortion, the best resolution that can be obtained by a normal, earth-based, visible-light telescope is about 0.3 arcsecond (there are 60 arcminutes in a degree and 60 arcseconds in an arcminute).Using Rayleigh's criterion, calculate the diameter of an earth-based telescope that gives this resolution with 700 nm light

Answer:

The diameter is  D = 0.59 \  m    

Explanation:

From the question we are told that

      The best resolution is  \theta  =  0.3 \  arcsecond

       The  wavelength is  \lambda  =  700 \  nm =  700 *10^{-9 } \  m

       

Generally the

         1 arcminute  = >  60 arcseconds

=>      x arcminute =>   0.3 arcsecond

So

       x =  \frac{0.3}{60 }

=>    x = 0.005 \  arcminutes

Now

         60 arcminutes  =>  1 degree

          0.005 arcminutes = >  z degrees  

=>       z =  \frac{0.005}{60 }

=>      z =  8.333 *10^{-5}  \ degree

Converting to radian  

           \theta  = z =  8.333 *10^{-5}  * 0.01745 = 1.454 *10^{-6} \  radian

Generally the resolution is mathematically represented as

            \theta  =  \frac{1.22 *  \lambda  }{ D}

=>    D =  \frac{1.22 * \lambda }{\theta }

=>     D =  \frac{1.22 * 700 *10^{-9} }{ 1.454 *10^{-6} }    

=>     D = 0.59 \  m    

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