Question

Mars, which has a radius of 3.4 × 106 m and a mass of 6.4 × 1023 kg, orbits the Sun, which has a mass of 2.0 × 1030 kg at a distance of 2.3 × 1011 m. Which is greater, the tangential speed of Mars’s rotation or revolution?

Answer 1

The correct answer is: Revolution (The tangential speed of Mars' revolution is greater than that of Mars' rotation).

Explanation:

Given data:

Radius-of-Mars = r =  $3.4 * 10^6$ m

Distance-between-Sun-and-Mars = d = $2.3 * 10^{11}$ m

Now let us first calculate the Mars' rotation:

One complete rotation of Mars = 2πr = 2π($3.4*10^6$) = $21.36 * 10^6$ m

Tangential speed of Mars' rotation (by taking 24hours 37 minutes in seconds (88620s)—as Mars takes that many seconds to complete one rotation) = $\frac{21.36 * 10^6m}{88620} = 241.02 \frac{m}{s}$

Now let's calculate the Mars' revolution (around Sun):

One complete revolution of Mars around Sun = 2πd = 2π($2.3 * 10^{11}$) = $1.45 * 10^{12}$ m

Tangential speed of Mars' revolution (by taking 687 days in seconds (59356800s)—since Mars takes 687days to complete one revolution)= $\frac{1.45 * 10^{12}m}{59356800s} = 24.42 * 10^{3}\frac{m}{s}$

So we can see that the tagential speed of:

Revolution > Rotation

$24.42 * 10^{3}\frac{m}{s}$ > $241.02\frac{m}{s}$

Hence, the correct answer is Revolution.

Answer 2

The correct answer is: revolution