    # Explain why values of gravitational potential near to an isolated mass are all negative.

About negative energies: they set no problem:

On this context, only energy differences have significance. Negative energy appears because when you've made the integration, you've set one point where you set your energy to 0. In this case, you have chosen that $PE_1 = 0$ for $r = \infty$. If you've set $PE_1 = 1000$ at $r = \infty$, the energy was positive for some r.

However, the minus sign is important, as it is telling you that the test particle is losing potential energy when moving to $r = 0$, this is true because it is accelerating, causing an increase in $KE$:

let's calculate the $\Delta PE_1$ for a particle moving in direction of $r = 0$: $r_i = 10$ and $r_f = 1$:

$\Delta PE_1 = PE_f - PE_i = Gm(-1 - (-0.1)) = -Gm\times0.9 < 0$

as expected: we lose $PE$ and win $KE$.

Second bullet: yes, you are right. However, it is only true IF they are point particles: has they normally have a definite radius, they collide when $r = r_1 + r_2$, causing an elastic or inelastic collision.

Third bullet: you are right with $PE_2 = mgh$, however, again, you are choosing a given referential: you are assuming $PE_2 = 0$ for $y = 0$, which, on the previous notation, means that you were setting $PE_1 = 0$ for $r = r_{earth}$.

The most important difference now is that you are saying that an increase in h is moving farther in r (if you are higher, you are farther from the Earth center).

By making the analogy to the previous problem, imagine you want to obtain the $\Delta PE_2$. In this case, you begin at $h_i = 10$ and you want move to $h_f = 1$ (moving in direction to Earth center, like $\Delta PE_1$:

$\Delta PE_2 = PE_{f} - PE_{i} = 1mg - 10mg = -9mg < 0$.

As expected, because we are falling, we are losing $PE$ and winning $KE$, the same result has $PE_1$

Fourth bullet: they both represent the same thing. The difference is that $gh$ is the first term in the Taylor series of the expansion of $PE_1$ near $r = r_{Earth}$. As exercise, try to expand $PE_1(r)$ in a taylor series, and show that the linear term is:

$PE_1 = a + \frac{Gm(r-r_{earth})}{r_{earth}^2}$.

Them numerically calculate $Gm/r_{earth}^2$ (remember that $m=m_{earth}$). If you haven't made this already, I guess you will be surprised.

So, from what I understood, your logic is totally correct, apart from two key points:

• energy is defined apart of a constant value.

• in the $PE_1$, increase r means decrease $1/r$, which means increase $PE_2 = -Gm/r$. In $PE_2$, increase h means increase $PE_2=mgh$.

Remember to keep the negative sign in your solution for gravitational potential. However, if you’re asked for the ‘change in’ gravitational potential, no negative sign should be included since you are finding a difference in values (between 0 at infinity and the gravitational potential from your calculation).

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There is a big difference between g and G (sometimes referred to as ‘little g’ and ‘big G’ respectively), g is the gravitational field strength and G is Newton’s gravitational constant. Make sure not to use these interchangeably!

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(a) Define gravitational potential.

(b) Explain why values of gravitational potential near to an isolated mass are all negative.

(c) Earth may be assumed to be an isolated sphere of radius 6.4 × 103 km with its mass of 6.0 × 1024 kg concentrated at its centre. An object is projected vertically from the surface of the Earth so that it reaches an altitude of 1.3 × 104 km.

Calculate, for this object,

(i) change in gravitational potential,

(ii) speed of projection from the Earth’s surface, assuming air resistance is negligible.

(d) Suggest why the equation

v2 = u2 + 2as

is not appropriate for calculation in (c)(ii).

Solution:

(a) Gravitational potential (at a point) is defined as the work done in bringing/moving unit mass from infinity to the point.

(b) The potential at infinity is defined as being zero. The forces are always attractive, so work got out in moving to point (work is done on the mass when moving it to infinity).

Gravitational potential, φ = - GM / R = - GM × (1/R)

{The distances should be converted in metre. Initial position is at the surface of the Earth, which is a distance of 6.4×106m from the centre of the Earth. The final distance is 1.3×107m (altitude = distance above surface) + 6.4×106m (distance of surface of the centre) = 1.94×107m}

Change in potential = (6.67×10-11) (6.0×1024) × ({6.4×106}-1 – {1.94×107}-1)

Change in potential = 4.19 × 107 J kg-1 (ignore sign)

The kinetic energy is converted to gravitational potential energy as the height of the object from the surface of the Earth increases.

v2 = 2 × 4.19×107 = 8.38 × 107

(d) The acceleration is not constant.

Reference: Past Exam Paper – June 2003 Paper 4 Q1 #### Page 2

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This is all down to how gravitational potential is defined. The gravitational potential at a point is defined as the work done per unit mass to move a mass from infinity to the point in the gravitational field. As all gravitational fields are attractive, this means work must always be done to move a mass AWAY from the centre of the gravitational field, but as gravitational potential concerns the work done per unit mass to move the mass TOWARDS the point in the gravitational field, it always has a negative value.

Question 892: [Gravitation]

(a) Define gravitational potential.

(b) Explain why values of gravitational potential near to an isolated mass are all negative.

(c) Earth may be assumed to be an isolated sphere of radius 6.4 × 103 km with its mass of 6.0 × 1024 kg concentrated at its centre. An object is projected vertically from the surface of the Earth so that it reaches an altitude of 1.3 × 104 km.

Calculate, for this object,

(i) change in gravitational potential,

(ii) speed of projection from the Earth’s surface, assuming air resistance is negligible.

(d) Suggest why the equation

v2 = u2 + 2as

is not appropriate for calculation in (c)(ii).

Reference: Past Exam Paper – June 2003 Paper 4 Q1

Solution 892:

(a) Gravitational potential (at a point) is defined as the work done in bringing/moving unit mass from infinity to the point.

(b) The potential at infinity is defined as being zero. The forces are always attractive, so work got out in moving to point (work is done on the mass when moving it to infinity).

(c)

(i)

Gravitational potential, φ = - GM / R = - GM × (1/R)

{The distances should be converted in metre. Initial position is at the surface of the Earth, which is a distance of 6.4×106m from the centre of the Earth. The final distance is 1.3×107m (altitude = distance above surface) + 6.4×106m (distance of surface of the centre) = 1.94×107m}

Change in potential = (6.67×10-11) (6.0×1024) × ({6.4×106}-1 – {1.94×107}-1)

Change in potential = 4.19 × 107 J kg-1 (ignore sign)

(ii)

The kinetic energy is converted to gravitational potential energy as the height of the object from the surface of the Earth increases.

½ mv2 = mΔφ

v2 = 2 × 4.19×107 = 8.38 × 107

Speed v = 9150 m s-1

(d) The acceleration is not constant.

Question 893: [Kinematics > Graph]

Graph shows how the velocity v of a firework rocket changes with time t.

At which point on the graph does the rocket have the greatest acceleration? Reference: Past Exam Paper – November 2013 Paper 13 Q7

Solution 893:

The gradient of a velocity-time graph gives the acceleration.

Thus, the acceleration is greatest where the value of the gradient is greatest. Visually, this is be identified by the steepness of the tangent at a point – the steeper the tangent, the greater is the gradient.

The tangent is steepest at point B, compared with the other points.

Question 894: [Dynamics]

When a golfer hits a ball his club is in contract with the ball for about 0.0005 s and the ball leaves the club with a speed of 70 m/s. The mass of the ball is 46 g. Determine the mean force on the ball?

Reference: ???

Solution 894:

From Newton’s 2nd law, Force F is equal / proportional to rate of change of momentum. (F = Δp / t)

Change in momentum is given by p = mΔv

(Assuming all other velocities are zero – during the contact, the ball is momentarily stationary)

So, Force = mΔv / t = (0.046 × 70) / 0.0005 = 6440N

Question 895: [Alternating Current > Rectification]

Alternating supply of frequency 50 Hz and having an output of 6.0 V r.m.s. is to be rectified so as to provide direct current for a resistor R. The circuit of Fig.1 is used. The diode is ideal. The Y-plates of a cathode-ray oscilloscope (c.r.o.) are connected between points A and B.

(a)

(i) Calculate maximum potential difference across the diode during one cycle.

(ii) State potential difference across R when the diode has maximum potential difference across it. Give a reason for your answer.

(b) Y-plate sensitivity of the c.r.o. is set at 2.0 V cm–1 and the time-base at 5.0 ms cm–1.

On Fig.2, draw the waveform that is seen on the screen of the c.r.o.

(c) A capacitor of capacitance 180 μF is connected into the circuit to provide smoothing of the potential difference across the resistor R.

(i) On Fig.1, show the position of the capacitor in the circuit.

(ii) Calculate energy stored in the fully-charged capacitor.

(iii) During discharge, potential difference across the capacitor falls to 0.43V0, where V0 is the maximum potential difference across the capacitor.

Calculate fraction of the total energy that remains in the capacitor after the discharge.

Reference: Past Exam Paper – November 2006 Paper 4 Q6

Solution 895:

(a)

(i) Peak voltage = 6√2 = 8.48 V

(ii) The potential difference is zero because EITHER there is no current in circuit (and V = IR) OR all p.d. is across the diode

(b)

The waveform is that of a half-wave rectification.

{Peak voltage = 8.48V. 1cm represents 2.0V. 8.48V is represented by 8.48 / 2.0 = 4.24cm}

The peak height is at about 4.25 cm

{Frequency f is 50Hz. Period T = 1/f = 1/50 = 0.02s = 20ms. Half-period = 10ms. 1cm represents 5.0ms. 20ms is represented by 20 / 5 = 4cm. Half-period is represented by 2cm.}

The half-period spacing is 2.0 cm (c)

(i) The capacitor should be shown in parallel with the resistor. (ii)

EITHER Energy = ½ CV2      OR = ½ QV and Q = CV

Energy = ½ × (180×10–6) × (6√2)2 = 6.48 × 10–3 J

(iii)

{EITHER Energy is proportional to V2. Fraction = remaining energy / total energy.

Remaining energy is proportional to (0.43V0)2. Total energy is proportional to V02.

So, fraction = (0.43V0)2 / V02 = 0.432

OR The actual value of the remaining energy could be calculated and the fraction could be found using the previously calculated value of the total energy in (c)(ii). But this is too time-consuming.}

EITHER fraction = 0.432        OR final energy = 1.2 mJ

Fraction = 0.18

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