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

Answer 2

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

Step-by-step 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) =
`(21.36 * 10^6m)/(88620) = 241.02 (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)=
`(1.45 * 10^(12)m)/(59356800s) = 24.42 * 10^(3)(m)/(s)`

So we can see that the tagential speed of:

Revolution > Rotation

`24.42 * 10^(3)(m)/(s)` >
`241.02(m)/(s)`

Hence, the correct answer is Revolution.