4) A proton moves from point ito point f, in the direction of a uniform electric

Question

4) A proton moves from point ito point f, in the direction of a uniform electric field. During this displacement: s paints) A the work done by the field is positive and the potential energy of the system increases. the done by the field is negative and the potential energy of the system increases. C. the work done by the field is positive and the potential energy of th system decreases. D. the work done by the field is negative and the potential energy of the system decreases. E. the work done by the field is positive and the potential energy of the system does not change. Skill exercise l (10 points) The large van DeGraff generator in Robert's office has a sphere with a radius of 0.6m. a) How much charge can be placed o the metal sphere before the electric field at the surface is 30x10" is enough to initiate a spark) b) What voltage is this relative to zero volts at infinity?

Explanation / Answer

Let the mass of the earth be me , mass of the sun be ms and the mass of the asteroid be ma

me = 5.972 x 1024 , ms = 1.989 x 1030 kg , ma = 5.1 x 1016 kg (the exponent appears to me as 16. even after zooming it is not clear. So, if it is not the case, don't worry. just change the value in the following calculation to correct it. or else comment below and I'll resolve the problem. )

Please note that, for simplicity we are not considering the presence of other planets or satellites in our solar system. Our system consists of only the aforementioned three bodies.

Now the Gravitational potential energy between two bodies at a distance r is given by

V = - G m1m2 / r where G is the universal gravitational constant and m1 and m2 are the masses of the two bodies. and r is the distance between them. (Note the -ve sign)

G = 6.67408 × 10-11 m3 kg-1 s-2

So, initially when the asteroid way away from the solar system, the distance r can be considered infinity from both earth and the sun leading to initial potential energy zero.

a.) When the collision takes place, the distance between the asteroid and the earth would be nothing but the radius of the earth which is 6371 km

So the potential energy between earth and the asteroid at the instant of collision would be Vea = - G mema / re

Substituting the values, we get,   Vea = - 6.67408 × 10-11 x 5.972 x 1024 x 5.1 x 10 16 / 6371000   ( make sure to use SI units everywhere )

Vea = - 3.189160502 x 1025  Joules (do not forget the negative sign)

Observe that the potential energy now is negative and initially, it was zero. So, the potential energy has been decreased. Yes, and this decreased potential energy is what is manifested as the Kinetic energy. (Remember, energy can neither be created nor destroyed. It only can be transferred). So, here the initial potential energy has turned into the kinetic energy.

Mathematically, energy can neither be created nor destroyed. So, KE + PE = 0 or KE = - PE

KE = - (PEfinal - PEinitial ) = - ( Vea - 0 ) = - Vea

Therefore increase in the kinetic energy of asteroid due to earth = KEea = - Vea = 3.189160502 x 1025   Joules

Please note that this Potential energy is not LOST, it got converted into Kinetic energy.

b.) Even in this case, use the same formula V = - G msma / r  

But here, r is the distance between the sun and the asteroid, which is nothing but the distance between the sun and the earth (since asteroid is colliding with the earth) which is 149.6 million km = 1.496 x 1011 m

therefore, Vsa = - 6.67408 × 10-11 x 1.989 x 1030 x 5.1 x 1016 / 1.496 x 1011

- Vsa = 6.770120011 x 1036 Joules

KEsa = - Vsa = 6.770120011 x 1036 Joules

This is the amount of potential energy that got converted in KE because of sun's pull.

c.) So the net Kinetic energy gain by the asteroid is the sum of changes in the Kinetic energy because of Sun and the Earth (Now, imagine how cumbersome it would have been, had we taken the effect of all the other planets and satellites into consideration. But actual astronauts do take a hell lot of more things into account !)

KEa = KEea + KEsa =  3.189160502 x 1025 + 6.770120011 x 1036 = 6.770120011 x 1036 Joules !!

LOL!!! Earth' s gravitational pull could hardly make any difference to that of SUN's. This is because of the massive weight of the sun as compared to the earth or any other body in our solar system. That is why our initial approximation (of neglecting the presence of other planets) holds well and good.

Now that we know the net gain in Kinetic energy of the asteroid, do not do the mistake of directly equating it to 0.5mavf2

Remember that, the asteroid, when was far away, had an initial speed of 6500 m/s

So the initial KE is KEinitial = 0.5 x ma x v i2 = 0.5 x 5.1 x 1016 x 65002 = 1.077375 x 1024 Joules

So, the final Kinetic Energy KEfin = KEinitial + KE = 1.077375 x 1024 + 6.770120011 x 1036 = 6.770120011 x 1036

Note that, the initial KE couldn't make any difference here, again owing to the Gigantic pull of the sun. But, in future problems, you may face a good initial velocity which can give a decent contribution to the final Kinetic Energy. So, do not forget this step.

Now, if vf is the final velocity, KEfinal = 0.5 x ma x vf2

6.770120011 x 1036 = 0.5 x 5.1 x 1016 x vf2

2.654949024 x 1020 = vf2

=> vf  = 1.629401431 x 1010  m/ s

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