Question

Estimate the kinetic energy of the Mars with respect to the Sun as the sum of the terms, that due to its daily rotation about its axis, and that due to its yearly revolution about the Sun. [Assume the Mars is a uniform sphere with mass = 6.4×10²³ kg, radius = 3.4×10⁶ m, rotation period 24.7 h, orbital period 686 d and is 2.3×10⁸ km from the Sun.]Please explain thoroughly.

Answer 1

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

Answer 2

The mass of  of 6.4 × 10²³ and periods of 24.7 h and 686 d gives the

kinetic energy of Mars as approximately 1.902 × 10³² J

How can the kinetic energy of Mars be calculated?

The rotational kinetic energy is given as follows;

$E_{rotational} = \mathbf{\dfrac{1}{2} \cdot I \cdot \omega^2}$

The moment of inertia of a sphere, I = $\mathbf{\frac{2}{5} \cdot M \cdot R^2}$

Where;

$\omega = \dfrac{2 \cdot \pi}{24.7 \times 60 \times 60} \approx 7.066 \times 10^{-5}$

Which gives;

$E_{rotational} = \dfrac{1}{2} \times \frac{2}{5} \times 6.4 \times 10^{23} \times \left(3.4 \times 10^6 \right)^2 \cdot \left(7.066 \times 10^{-5} \right)^2 \approx \mathbf{ 7.388 \times 10^{27}}$

Distance travelled in an orbit = 2 × π × 2.3 × 10⁸ = 1445132620.25 km

The translational speed, v, is therefore;

$v = \dfrac{1445132620.25 \times 10^3 }{686 \times 24 \times 60 \times 60} = \mathbf{24382.03}$

The translational speed, v ≈ 24382.03 m/s

The translational kinetic energy is therefore;

K.E. = 0.5 × 6.4 × 10²³ × 24382.03² = 190234.67 × 10²⁷

The kinetic energy of Mars with respect to the Sun is therefore;

$K.E._{total}$ = 190234.67 × 10²⁷ + 7.388 × 10²⁷ ≈ 1.902 × 10³²

Therefore;

• The total kinetic energy of Mars w.r.t the Sun ≈ 1.902 × 10³² J = $\underline{1.902 \times 10^8 \, YJ}$

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