Gravitational Potential Energy

[Taniaz]

Homework Statement

A body of mass m is taken at a constant speed from the surface of the Earth (radius = Re) to infinity.
(a) What is the work W1 done on the body in the process?
(b) If m is taken from a distance r > Re to infinity, how much work W2 is required?
(c)Which quantity is larger, W1 or W2? Why?
(d) How much work is required to take the body from Re to r?
(e) The body is dropped from r with no initial velocity. Find an expression for the velocity with which it hits the Earth's surface. (Neglect all frictional forces)
(f) With what minimum velocity must m be thrown upward in order to reach an altitude h = r-Re
(g) If m is thrown with the above velocity upward but not exactly vertically, will it reach the above altitude? Explain your answer.
(h) What is the potential energy U of the body at altitude h?
(i) What is the potential energy U of the body at the surface of the earth?
(j) Which is more, the potential energy at the surface or that at altitude h?How much more?
(k) Is the work done on m taking it from the surface to altitude h equal to the increase in potential energy of m?

Homework Equations

U(r) = -GMm/r

The Attempt at a Solution

For a) I did an integral from Re to infinity and got W1=GMm/R
For b) I did the same from r to infinity and got W2= GMm/r
For c) I took did GMm/r - GMm/Re = GMm (1/r - 1/Re) so W1>W2
For d) I took the integral from Re to r and got GMm ( -1/r + 1/Re)
I'm not sure of (e) onwards
For e) I have an idea that K.E = P.E so 1/2 (mv^2) = U but I'm not sure what U in this case will be. I know it can't be negative.

 

[Taniaz]

Addendum to my attempted solution:
I think for part h) it should just be U = -GMm/ (Re +h)
and part i) should just be either U=mgh or -GMm/Re
for j) we subtract the 2 and see I presume

for e) I believe the equation we'll be using is K1 + U1 = K2 + U2
K1 will be 0 because and U1 will be -GMm/r and K2 is 1/2 mv^2 and U2 is -GMm/Re?

And we do the same for f but K2 would be zero in this case?

 

[SammyS]

1. A body of mass m is taken at a constant speed from the surface of the Earth (radius = Re) to infinity.
(a) What is the work W1 done on the body in the process?
(b) If m is taken from a distance r > Re to infinity, how much work W2 is required?
(c)Which quantity is larger, W1 or W2? Why?
(d) How much work is required to take the body from Re to r?
(e) The body is dropped from r with no initial velocity. Find an expression for the velocity with which it hits the Earth's surface. (Neglect all frictional forces)
(f) With what minimum velocity must m be thrown upward in order to reach an altitude h = r-Re
(g) If m is thrown with the above velocity upward but not exactly vertically, will it reach the above altitude? Explain your answer.
(h) What is the potential energy U of the body at altitude h?
(i) What is the potential energy U of the body at the surface of the earth?
(j) Which is more, the potential energy at the surface or that at altitude h?How much more?
(k) Is the work done on m taking it from the surface to altitude h equal to the increase in potential energy of m?

2. U(r) = -GMm/r

3. For a) I did an integral from Re to infinity and got W1=GMm/R
For b) I did the same from r to infinity and got W2= GMm/r
For c) I took did GMm/r - GMm/Re = GMm (1/r - 1/Re) so W1>W2
For d) I took the integral from Re to r and got GMm ( -1/r + 1/Re)
I'm not sure of (e) onwards
For e) I have an idea that K.E = P.E so 1/2 (mv^2) = U but I'm not sure what U in this case will be. I know it can't be negative.

You don't actually state what quantities you are integrating

Your answer for part c) is correct, but not very convincing. What you have there is actually W₂ - W₁ which is negative if W₁ > W₂ . To show that 1/r - 1/R_e is negative, write it as a single fraction with a common denominator.

Your answers to parts a) through d) are correct.

For e): It's more proper to say that the increase in KE is equal to the decrease in PE .

You say that U (potential energy) can't be negative. However, in this case we usually consider U to be zero at infinite distance and that is the case where U is maximum. Therefore, U is negative at any closer distance. Also, notice your formula for U.
U(r) = -GMm/r​

Addendum to my attempted solution:
I think for part h) it should just be U = -GMm/ (Re +h)
and part i) should just be either U=mgh or -GMm/Re
for j) we subtract the 2 and see I presume

for e) I believe the equation we'll be using is K1 + U1 = K2 + U2
K1 will be 0 because and U1 will be -GMm/r and K2 is 1/2 mv^2 and U2 is -GMm/Re?

And we do the same for f but K2 would be zero in this case?

h) is incorrect . That's not how to subtract fractions - if that's what you intended.

Part i): U = mgh is good for potential energy near Earth's surface with U = 0 at the surface. - not good for this problem.

Part j): Use the basic idea behind potential energy .See if subtracting the two values is consistent with this.

 

[Taniaz]

Thank you for your reply.

Why do we have to subtract fractions for part h? It just asks for the potential and so substituting for r in
U(r)=-GM/r as (Re+h)? Or no, actually I think it's the potential at r - potential at Re.

And how would you do part f? Similar way to part e?

For part e) I get (-GMm/r) = 1/2 mv^2 but that doesn't seem right because you can't take the root of a negative number? Or would I need to take into account the potential at Re as well
So -GMm/r = 1/2 mv^2 + (-GMm/Re)? but even then -GMm/Re + GMm/r is still negative because r > Re and so - 1/Re + 1/r will be negative?

 

[SammyS]

Thank you for your reply.

Why do we have to subtract fractions for part h? It just asks for the potential and so substituting for r in
U(r)=-GM/r as (Re+h)?

You were correct. I had a brain cramp.

And how would you do part f? Similar way to part e?
For part e) I get (-GMm/r) = 1/2 mv^2 but that doesn't seem right because you can't take the root of a negative number? Or would I need to take into account the potential at Re as well
So -GMm/r = 1/2 mv^2 + (-GM/Re)? but even then -GM/Re + GMm/r is still negative?

It's the change in potential energy that you need for part e). (Be careful of signs.)

 

[Taniaz]

If we do it as K1+U1=K2+U2 or take it as increase in kinetic energy= decrease in potential energy we still get
1/2 mv^2 = (-GMm/Re) - (-GMm/r) which means the RHS is negative-still confused about this.

 

[Taniaz]

Ok no, I get it now for part e. Initially the velocity was 0 so the KE at r is 0.
So if we use K1 + U1 = K2 +U2 where 1 is at r and 2 is at Re we get
0 - GMm/r = 1/2 mv^2 - (GMm/Re)
So -GMm/r + GMm/Re =1/2 mv^2
GMm ( -1/r + 1/Re) = 1/2 mv^2
Now -1/r +1/Re is positive
Is that correct?

As decided for h it will be U=-GMm/(Re+h) at altitude and i) U=GMm/Re for the surface
Then for j we can just look at the inverse proportionality between U and the distance. As you increase the distance, U increases as well (presence of negative sign). So it's greater with Re+h? To find how much, we can just subtract the two?
k) is yes because the work done =change in energyJust not sure of f and g now.

 

[haruspex]

Ok no, I get it now for part e. Initially the velocity was 0 so the KE at r is 0.
So if we use K1 + U1 = K2 +U2 where 1 is at r and 2 is at Re we get
0 - GMm/r = 1/2 mv^2 - (GMm/Re)
So -GMm/r + GMm/Re =1/2 mv^2
GMm ( -1/r + 1/Re) = 1/2 mv^2
Now -1/r +1/Re is positive
Is that correct?

As decided for h it will be U=-GMm/(Re+h) at altitude and i) U=GMm/Re for the surface
Then for j we can just look at the inverse proportionality between U and the distance. As you increase the distance, U increases as well (presence of negative sign). So it's greater with Re+h? To find how much, we can just subtract the two?
k) is yes because the work done =change in energyJust not sure of f and g now.

That all looks right.
For f, if it is thrown up with the minimum velocity to reach height h, what will its velocity be when it gets there?

 

[Taniaz]

It will be 0 so k2 will be zero, our equation will then be
K1+U1=K2+U2
1/2 mv^2 + (-GMm/Re) = 0 + (-GMm/r)?

 

[haruspex]

It will be 0 so k2 will be zero, our equation will then be
K1+U1=K2+U2
1/2 mv^2 + (-GMm/Re) = 0 + (-GMm/r)?

Right.
What about g? Velocity is a vector, so what else can you deduce from the fact that velocity will be 0 at height h?

 

[Taniaz]

g will decrease?

Sorry the equation will be 1/2 mv^2 + (-GMm/Re)= 0 + (-GMm/(r-Re)
since h=r-Re?

 

[haruspex]

g will decrease?

No, I meant question g.

Sorry the equation will be 1/2 mv^2 + (-GMm/Re)= 0 + (-GMm/(r-Re)

No, you were right in post #9, but you can cancel a common factor.

 

[Taniaz]

But they're asking for U at h and not r?
It's going to break up into components because it's not exactly vertical now?

 

[haruspex]

But they're asking for U at h and not r?

Yes, they want the answer in terms of h, not in terms of r, but your equation in post #9 is correct and your equation in post #11 is wrong. Have another go at converting the post#9 equation from using r to using h.

It's going to break up into components because it's not exactly vertical now?

Yes, but that's not quite what I was hinting at. If you know the velocity as a vector is zero at h, what does that tell you about those components?

 

[Taniaz]

Ohhh sorry! My bad, if h=r-Re then r=h+Re so the denominator of the r fraction will be h+Re

So I got
1/2 mv^2 + (-GMm/Re) = 0 + (-GMm/r)
1/2 mv^2 = (-GMm/r) + (GMm/Re)
1/2 mv^2 = GMm ( -1/r + 1/Re)
1/2 v^2 = GM ( -Re + r / rRe) but h=r-Re so
v^2= 2GM (h /rRe)
and v is the square root of that.

So vx^2 + vy^2=0?

Components equal and opposite to one another?

 

[haruspex]

denominator of the r fraction will be h+Re

Yes.

1/2 mv^2 = GMm ( -1/r + 1/Re)

Yes.

v^2= 2GM (h /rRe)

No.

So vx^2 + vy^2=0?

Yes.

Components equal and opposite to one another?

No.

 

[Taniaz]

Ok, I get it.
I'll have to substitute for r=Re+h in (-1/r + 1/Re)

-1/(h+Re) + 1/Re = -Re+Re+h / Re (Re+h) = h / Re(Re+h)
So v^2 = 2GM ( h / Re(Re+h))

Components=0

 

[haruspex]

Ok, I get it.
I'll have to substitute for r=Re+h in (-1/r + 1/Re)

-1/(h+Re) + 1/Re = -Re+Re+h / Re (Re+h) = h / Re(Re+h)
So v^2 = 2GM ( h / Re(Re+h))

Components=0

Yes, and yes.

 

[Taniaz]

Yes!
So for g it won't be able to reach this altitude then?

 

[haruspex]

Yes!
So for g it won't be able to reach this altitude then?

That's right, though to complete the argument you need to say why the horizontal component will not be zero at the top of the trajectory.

 

[Taniaz]

The horizontal component will not be zero at the top?

 

[haruspex]

The horizontal component will not be zero at the top?

That's right, but why?

 

[Taniaz]

But we just said both components were zero? That's how v will be zero?

 

[haruspex]

But we just said both components were zero? That's how v will be zero?

Let's try another tack.
You found the initial velocity needed to reach height h if it is thrown up vertically. If it is thrown up with that speed but at angle theta to the vertical, what height will it reach?

 

[Taniaz]

Well for a projectile it would be y = (vo)t sin (theta) - 1/2 gt^2 but theta is the angle to the horizontal

 

[haruspex]

Well for a projectile it would be y = (vo)t sin (theta) - 1/2 gt^2 but theta is the angle to the horizontal

Take theta as relative to the horizontal if you are more comfortable with that. But that equation tells you the height at time t, not the maximum height.

 

[Taniaz]

Oh then it's h = vo^2 sin^2(theta) / 2g

 

[haruspex]

Oh then it's h = vo^2 sin^2(theta) / 2g

That would be right for constant gravity, g. But this question considers the more accurate treatment, where gravity drops as height increases. You need to use an equation you found earlier, but modify it by incorporating the launch angle.

 

[Taniaz]

The equation for minimum velocity into h?
v = sqrt (2GM (h / Re(Re+h))
So vo^2 would be just 2GM (h / Re(Re+h)

 

[haruspex]

The equation for minimum velocity into h?
v = sqrt (2GM (h / Re(Re+h))
So vo^2 would be just 2GM (h / Re(Re+h)

Right, and you arrived at that by energy arguments. Now suppose we throw it up with speed vo but at angle theta to the horizontal. What will be its horizontal velocity component when at maximum height, and why?

 

[Taniaz]

My brain is dead, I don't know...

 

[Biker]

Lets remember what we studied about projectile motion. If the only force is gravity then the horizontal velocity vector is always constant.
and what do we know about the vertical velocity at maximum height?

So you started with ... velocity and ended up with ... velocity (Fill the dots)

So is this loss in kinetic energy is equal to the gain in potential energy? If yes then they will reach it if not then rip astronauts

 

[Taniaz]

Vertical velocity at maximum height is 0.

 

[Biker]

Vertical velocity at maximum height is 0.

and the horizontal? as we know it doesn't change? So if we can figure out what it is at the beginning it is the same at the end. So what is the horizontal component?

 

[Taniaz]

vx=vo cos(theta)?