Answer:
K = 2.07 10³⁹ J
Explanation:
This problem must be solved using rotational kinematics.
Kinetic energy has the form
K = ½ m v²
Linear velocity is related to angular velocity.
v = w r
replace
K = ½ m R² w²
With this equation we can find the total kinetic energy of Mars, formed by the rotation energy plus the translational energy
K = Kr + Kt
Where Kr is the energy by the rotation on its axis and Kt is the energy by the rotation around the sun.
Let's reduce to SI units
Rotation T₁ = 24.7 h (3600 s / 1 h) = 88920 s
R₁ = 3.4 10⁶ m
translation T₂ = 686 day (24 h / 1 day) (3600 s / 1 h) = 5.927 10⁷ s
R₂ = 2.3 10⁸ km (1000 m / 1 km) = 2.3 10¹¹ m
The angular velocity is the angle rotated (radians) between the time taken, in this case as the order gives us the angle is 2pi rad (360º). Remember that the equations work only in radians
Rotation
wr = 2π / T₁
wr = 2 π / 88920
wr = 7.066 10⁻⁵ rad / s
Translation
wt = 2 π / T₂
wt = 2 π / 5,927 10⁷
wt = 1.06 10⁻⁷-7 rad/s
Let's explicitly write the equation of kinetic energy and calculate
K = ½ m R₁² wr² + ½ m R₂² wt²2
K = ½ m (R₁² wr² + R₂² wt²)
K = ½ 6.4 10²³ [(3.4 10⁶)² 7.066² + (2.3 10¹¹)² (1.06 10⁻⁷)²]
K = 3.2 10²³ [61.49 10¹² + 6.414 10¹⁵]
K = 3.2 10²³ [61.49 10¹² + 6414 10¹²]
K = 3.2 10²³ (6475 10¹²)
K = 2.07 10³⁹ J
