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Mars, which has a radius of [tex]$3.4 \times 10^6 \, m$[/tex] and a mass of [tex]$6.4 \times 10^{23} \, kg$[/tex], orbits the Sun, which has a mass of [tex]$2.0 \times 10^{30} \, kg$[/tex], at a distance of [tex]$2.3 \times 10^{11} \, m$[/tex].

- Mars's revolution around the Sun is 687 days.
- It takes 24 hours and 37 minutes to complete one rotation.

Which is greater, the tangential speed of Mars's rotation or revolution? [tex]$\square$[/tex]



Answer :

GinnyAnswer
To determine which is greater, the tangential speed of Mars's rotation or its revolution around the Sun, we need to calculate both speeds separately using given data.

### Calculate Tangential Speed of Mars's Revolution

1. Convert Revolution Period to Seconds:
- Mars takes 687 days to orbit the Sun.
- [tex]\( 1 \text{ day} = 24 \text{ hours} = 24 \times 60 \times 60 \text{ seconds} \)[/tex]
- Total revolution time [tex]\( T_{\text{revolution}} \)[/tex] in seconds:
[tex]\[ T_{\text{revolution}} = 687 \times 24 \times 60 \times 60 \][/tex]

2. Calculate Tangential Speed for Revolution:
- Distance from the Sun, [tex]\( d = 2.3 \times 10^{11} \text{ meters} \)[/tex]
- Tangential speed [tex]\( v_{\text{revolution}} \)[/tex]:
[tex]\[ v_{\text{revolution}} = \frac{2 \pi d}{T_{\text{revolution}}} \][/tex]

### Calculate Tangential Speed of Mars's Rotation

1. Convert Rotational Period to Seconds:
- Mars's rotational period is 24 hours and 37 minutes.
- [tex]\( 1 \text{ hour} = 60 \text{ minutes} \)[/tex]
- Total rotational period [tex]\( T_{\text{rotation}} \)[/tex] in hours:
[tex]\[ T_{\text{rotation}} = 24 + \frac{37}{60} \text{ hours} \][/tex]
- Convert this to seconds:
[tex]\[ T_{\text{rotation}} \text{ (seconds)} = T_{\text{rotation}} \times 60 \times 60 \][/tex]

2. Calculate Tangential Speed for Rotation:
- Radius of Mars, [tex]\( r_{\text{Mars}} = 3.4 \times 10^6 \text{ meters} \)[/tex]
- Tangential speed [tex]\( v_{\text{rotation}} \)[/tex]:
[tex]\[ v_{\text{rotation}} = \frac{2 \pi r_{\text{Mars}}}{T_{\text{rotation}}} \][/tex]

### Comparison of the Speeds

From the above calculations, we determine:
- The tangential speed of Mars's revolution around the Sun is approximately [tex]\( 24346.54 \text{ m/s} \)[/tex].
- The tangential speed of Mars's rotation is approximately [tex]\( 241.06 \text{ m/s} \)[/tex].

### Conclusion

The tangential speed of Mars's revolution around the Sun ([tex]\( 24346.54 \text{ m/s} \)[/tex]) is significantly greater than the tangential speed of Mars's rotation ([tex]\( 241.06 \text{ m/s} \)[/tex]). Therefore, the tangential speed of Mars's revolution is greater. [tex]\( \boxed{} \)[/tex]

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