Question Video: Finding the Surface Gravity of Mars as a Ratio of the Surface Gravity of Earth Physics • 9th Grade

Mars has a mass of 6.42 × 10²³ kg and a radius of 3,390 km. What is the surface gravity of Mars as a ratio of the surface gravity of Earth? Use a value of 9.81 m/s² for the surface gravity of Earth. Give your answer to two decimal places.

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Video Transcript

Mars has a mass of 6.42 times 10 to the 23 kilograms and a radius of 3,390 kilometers. What is the surface gravity of Mars as a ratio of the surface gravity of Earth? Use a value of 9.81 meters per second squared for the surface gravity of Earth. Give your answer to two decimal places.

Now, this question is about comparing the surface gravities on Mars and Earth. So we need to recall the equation for surface gravity. That is, 𝑎 is equal to 𝐺𝑀 over 𝑟 squared, where 𝑎 is the surface gravity. And for Earth, this is given to us as 9.81 meters per second squared. 𝐺 is the universal gravitational constant, which is equal to 6.67 times 10 to the minus 11 meters cubed per kilogram second squared. 𝑀 is equal to the mass of the planet, which we’re given for Mars as 6.42 times 10 to the 23 kilograms. And 𝑟 is the radius of the planet, which we’re given for Mars as 3,390 kilometers.

So let’s start by working out the surface gravity on Mars. We have surface gravity on Mars is equal to the universal gravitational constant 𝐺 times the mass of Mars divided by the radius of Mars squared, which is equal to 6.67 times 10 to the minus 11 meters cubed per kilogram second squared times 6.42 times 10 to the 23 kilograms divided by 3,390 kilometers squared.

Now, notice that we have kilometers in the denominator and meters in the numerator. We need to convert these to the same units, which we do by recalling that one kilometer is equal to 1,000 meters. And therefore, 3,390 kilometers is equal to 3,390,000 meters.

Having substituted that value into the equation, we can now solve it. And we find that the surface gravity on Mars is equal to 3.726 meters per second squared. Now we need to express this as a ratio of the surface gravity of Earth. This means finding the surface gravity on Mars divided by the surface gravity on Earth. And that is equal to 3.726 meters per second squared divided by 9.81 meters per second squared. This comes to about 0.3798.

And notice that we had units of meters per second squared on both the numerator and the denominator. That means these cancel out, and what we’re left with has no units. We call this a dimensionless value. And we need to express this to two decimal places. So that comes to 0.38, which means the surface gravity on Mars as a ratio of the surface gravity on Earth is 0.38.

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