A robotic rover on Mars finds a spherical rock with a diameter of 10 centimeters [cm]. The rover picks up the rock and lifts it 20 centimeters [cm] straight up. The resulting potential energy of the rock relative to the surface is 2 joules [J]. Gravitational acceleration on Mars is 3.7 meters per second squared [m/s2]. What is the specific gravity of the rock?

The specific gravity of a solid [tex]SG[/tex] (also called relative density) is the ratio of the density of that solid [tex]\rho_{rock}[/tex] to the density of water [tex]\rho_{water}=1 kg/m^{3}[/tex] (normally at [tex]4\°C[/tex]):

[tex]SG=\frac{\rho_{rock}}{\rho_{water}}[/tex] (1)

On the other hand, the density of the rock is calculated by:

[tex]\rho_{rock}=\frac{m_{rock}}{V_{rock}}[/tex] (2)

Where:

[tex]m_{rock}[/tex] is the mass of the rock

[tex]V_{rock}=\frac{4}{3} \pi r^{3}[/tex] is the volume of the rock, since is spherical

Well, we already know the value of [tex]\rho_{water}[/tex], but we need to find [tex]\rho_{rock}[/tex] in order to find the rock's specific gravity; and in order to do this, we firsly have to find [tex]m_{rock}[/tex] and then calculate [tex]V_{rock}[/tex]:

In the case of the mass of the rock, we can calclate it by the following equation:

[tex]W_{rock}=m_{rock}g_{mars}[/tex] (3)

Where:

[tex]W_{rock}[/tex] is the weight if the rock in mars

[tex]g_{mars}=3.7 m/s^{2}[/tex] is the acceleration due gravity in Mars

Isolating [tex]m_{rock}[/tex]:

[tex]m_{rock}=\frac{W_{rock}}{g_{mars}}[/tex] (4)

[tex]m_{rock}=\frac{W_{rock}}{3.7 m/s^{2}}[/tex] (5)

To find [tex]W_{rock}[/tex] we can use the following equation of the potential gravitational energy [tex]U[/tex]:

[tex]U=W_{rock}H[/tex] (6)

Where:

[tex]U=2 J=2 Nm[/tex] is the potential energy

[tex]H=20 cm \frac{1m}{100 cm}=0.2 m[/tex] is the height at which the rock has the mentioned potential energy

Isolating [tex]W_{rock}[/tex]:

[tex]W_{rock}=\frac{U}{H}[/tex] (7)

[tex]W_{rock}=\frac{2 Nm}{0.2 m}[/tex] (8)

[tex]W_{rock}=10 N[/tex] (9)

Substituting (9) in (5):

[tex]m_{rock}=\frac{10 N}{3.7 m/s^{2}}[/tex] (10)

[tex]m_{rock}=2.702 kg[/tex] (11)

Substituting (11) in (2):

[tex]\rho_{rock}=\frac{2.702 kg}{V_{rock}}[/tex] (12) At this point we only need to find the volume of the rock, knowing its diameter is [tex]d=10 cm[/tex], hence its radius is [tex]r=\frac{d}{2}=5 cm[/tex]

[tex]V_{rock}=\frac{4}{3} \pi (5 cm)^{3}[/tex] (13)

[tex]V_{rock}=523.59 cm^{3} \frac{1 m^{3}}{(100 cm)^{3}}=0.000523 m^{3}[/tex] (14)

Substituting (14) in (12):

[tex]\rho_{rock}=\frac{2.702 kg}{0.000523 m^{3}}[/tex] (15)

[tex]\rho_{rock}=5166.34 kg/m^{3}[/tex] (16)

Substituting (16) in (1):

[tex]SG=\frac{5166.34 kg/m^{3}}{1 kg/m^{3}}[/tex] (17)

Finally we obtain the specific gravity of the rock:

[tex]SG=5166.347[/tex]