Two objects have masses of #3 MG# and #15 MG#. How much does the gravitational potential energy between the objects change if the distance between them changes from #130 m# to #24 m#?

Question

Two objects have masses of #3   MG# and #15  MG#. How much does the gravitational potential energy between the objects change if the distance between them changes from #130   m# to #24  m#?

Nora Arzate · Accepted Answer

The change in gravitational potential energy is #=10.2*10^-5J#

The potential energy per kilogram at a point in a field is known as the gravitational potential.

So the units are #J, "Joules"#

#Phi=-G(M_1M_2)/R#

The universal constant of gravitation is

G is 6.67 * 10^-11 Nm^2 kg^-2.

The masses causing the field is #=M_1 kg# and #=M_2 kg#

The mass is #M_1=3MG=3*10^6g=3*10^3kg#

The mass is #M_2=15MG=15*10^6g=15*10^3kg#

The distance between the centers is #=Rm#

The distance #R_1=130m#

The distance #R_2=24m#

Consequently,

#Phi_1=(-G*(3*10^3*15*10^3)/130)#

#Phi_2=(-G*(3*10^3*15*10^3)/24)#

Thus,

#Phi_1-Phi_2=(-G*(3*10^3*15*10^3)/130)-(-G*(3*10^3*15*10^3)/24)#

#=3*15*10^6*6.67*10^-11(1/24-1/130)#

#=10.2*10^-5J#

Lily Adams · Answer

undefined The gravitational potential energy between the objects changes by approximately -3.08 × 10^14 joules.

Nicholas Agrusa · Answer

undefined The change in gravitational potential energy between the two objects can be calculated using the formula for gravitational potential energy:

$$ \Delta U = -\frac{G \cdot m_1 \cdot m_2}{r_2} + \frac{G \cdot m_1 \cdot m_2}{r_1} $$

Where:
- $ \Delta U $ is the change in gravitational potential energy.
- $ G $ is the gravitational constant, approximately $ 6.674 \times 10^{-11} \, \text{m}^3/\text{kg}\cdot\text{s}^2 $.
- $ m_1 $ and $ m_2 $ are the masses of the two objects in kilograms.
- $ r_1 $ and $ r_2 $ are the initial and final distances between the objects in meters.

Given:
$ m_1 = 3 \, \text{MG} = 3 \times 10^6 \, \text{kg} $,
$ m_2 = 15 \, \text{MG} = 15 \times 10^6 \, \text{kg} $,
$ r_1 = 130 \, \text{m} $, and
$ r_2 = 24 \, \text{m} $.

Substitute the values into the formula and calculate:

$$ \Delta U = -\frac{6.674 \times 10^{-11} \times 3 \times 10^6 \times 15 \times 10^6}{24} + \frac{6.674 \times 10^{-11} \times 3 \times 10^6 \times 15 \times 10^6}{130} $$

$$ \Delta U \approx -\frac{3.0033 \times 10^5}{24} + \frac{3.0033 \times 10^5}{130} $$

$$ \Delta U \approx -12513750 + 57947.308 $$

$$ \Delta U \approx -12455802.692 \, \text{Joules} $$

Therefore, the change in gravitational potential energy between the objects is approximately $ -1.255 \times 10^7 \, \text{Joules} $.