Question 9
(a) By reference to the definition of gravitational potential, explain why
gravitational potential is a negative quantity. [2]
(b) Two stars A and B have their surfaces separated by a distance of 1.4 ×
1012 m, as illustrated in Fig. 1.1.
Fig. 1.1
Point P lies on the line joining the centres of the two stars. The
distance x of point P from the surface of star A may be varied.
The variation with distance x of the gravitational potential φ at point
P is shown in Fig. 1.2.
Fig. 1.2
A rock of mass 180 kg moves along the line joining the centres of the
two stars, from star A towards star B.
(i) Use data from Fig. 1.2 to calculate the change in kinetic energy of
the rock when it moves from the point where x = 0.1 × 1012 m to the
point where x = 1.2 × 1012 m.
State whether this change is an increase or a decrease. [3]
(ii) At a point where x = 0.1 × 1012 m, the speed of the rock
is v.
Determine the minimum speed v such that the rock reaches the point where
x = 1.2 × 1012 m. [3]
[Total: 8]
Reference: Past Exam Paper – June 2016 Paper 41 & 43 Q1
Solution:
(a) The (gravitational) potential at infinity is
defined as/is zero.
The (gravitational) force is attractive so work
got out/is done as the object moves from infinity. (so the potential is
negative)
(b)
(i)
{Gravitational potential is defined to be zero at
infinity (and this is the maximum value).
The more negative the value, the smaller is the
gravitational potential.
From the graph, it is observed that the gravitational
potential is more negative at x = 1.2 × 1012 m (at B) compared to
what it is at A. Thus, from A to B, there is an overall decrease in
gravitational potential. This would appear as an increase in KE of the rock.}
{Change in potential energy: ΔE = mΔφ
The change in KE is equal to the change in PE.
From the graph,
At x = 1.2 × 1012 m, φ = – 14 × 108 J kg-1
At x = 0.1 × 1012 m, φ = – 10 × 108 J kg-1
Δφ = final φ – initial φ = – 14 × 108 – (– 10 × 108) = – 4 × 108 J kg-1
Below, only the magnitude is being considered as we
have already determined that the KE is gained.
ΔE = mΔφ}
{Note that the equation ΔE = mΔφ = - GMm / r is valid
for a body in the field of a single planet/starbody. When there are 2 stars
(as in this case), we need to refer to the graph to know how the potential
varies with position.}
ΔE = 180 × (14 – 10) × 108
ΔE = 7.2 × 1010 J
The change is an increase {in KE}.
(ii)
{The rock needs to reach the point where the field
is zero. In the graph, this is where the gradient is zero.
Value of φ = – 4.4 × 108 J kg-1
The rock needs to reach only this point. The rock
would be attracted by the field of the other planet afterwards.
ΔE = mΔφ}
Energy required = 180 × (10 – 4.4) × 108
{The KE is equal to this amount of energy.
½ mv2 = 180 × (10 – 4.4) × 108 }
½ × 180 × v2 = 180 × (10 – 4.4) × 108
Minimum speed v = 3.3 × 104 m s–1
