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Ganymede is the largest moon in the solar system, with a mass of 1.48 times 10 to the power of 23 kilograms and a radius of 2,630 kilometers.
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What is the surface gravity on Ganymede?
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Give your answer to two decimal places.
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In this question, weโre looking at Ganymede, Jupiterโs largest moon and in fact the largest moon in the solar system.
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Weโve been asked to work out its surface gravity.
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And surface gravity is the acceleration that one object will experience due to the gravitational force between it and another object when it is on the surface of the other object.
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In this case, the object whose surface weโre considering is Ganymede, and the other object can be anything.
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For example, it could be an astronaut.
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To work out the acceleration that our astronaut would experience.
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We will first work out the gravitational force acting between the astronaut and the moon, which we will call ๐น.
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We can do this using Newtonโs law of gravitation.
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Newtonโs law of gravitation tells us that the force acting between two massive objects, in our case Ganymede and the astronaut, is equal to capital ๐บ, the gravitational constant, multiplied by the product of the two masses divided by the squared distance between the objectsโ center of masses.
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Applying this to our problem, we will call the mass of Ganymede capital ๐, the mass of our astronaut lowercase ๐, and the distance between the center of mass of Ganymede and the center of mass of the astronaut, we know is the radius of Ganymede, which we will call ๐.
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We know this because the astronaut is on the surface of Ganymede.
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Now that we have an expression for the force acting between Ganymede and the astronaut, we can use Newtonโs second law of motion to relate this force to the acceleration that the astronaut will experience.
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Newtonโs second law tells us that the force acting on the astronaut is equal to the mass of the astronaut multiplied by the acceleration the astronaut would experience.
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And we can rearrange this to make acceleration the subject of the equation.
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We will divide both sides by the mass of the astronaut ๐, where we see that the ๐ in the numerator and denominator on the right will cancel, leaving us with just an expression for the acceleration of the astronaut.
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Rewriting this so acceleration is on the left, we can now substitute our expression for ๐น that we calculated using Newtonโs law of gravitation into this equation, where we see that acceleration is equal to big ๐บ multiplied by capital ๐ multiplied by lowercase ๐ divided by ๐ squared all divided by ๐.
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And here, we can see that the mass of the astronaut, lowercase ๐, will cancel in the numerator and the denominator of this fraction.
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And this leaves us with just a one in the denominator.
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So we can actually write this fraction as just its numerator.
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So ๐ is equal to big ๐บ multiplied by capital ๐ divided by ๐ squared.
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In our specific question, the acceleration that the astronaut will experience on the surface of Ganymede is equal to big ๐บ multiplied by the mass of Ganymede divided by the radius of Ganymede squared.
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All we have to do now is get values for ๐บ, capital ๐, and ๐ and substitute them into this equation to calculate the surface gravity on Ganymede.
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๐บ is the gravitational constant and has a value of 6.67 times 10 to the power of negative 11 meters cubed per kilogram-seconds squared.
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The question tells us that Ganymede has a mass of 1.48 times 10 to the power of 23 kilograms.
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And we are also told that the radius of Ganymede is 2,630 kilometers.
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Before we go any further, we must check our units.
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In the numerator of our fraction, we have ๐บ multiplied by ๐, which will give as units of meters cubed per kilogram-seconds squared multiplied by kilograms in the numerator.
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And we see that the kilograms in the denominator of this fraction on the left and the kilograms on the right will cancel.
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So our units in the numerator actually simplify to meters cubed per seconds squared.
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In the denominator, we have ๐ squared, which gives us units of kilometers squared.
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And here we run into a problem.
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In the numerator, we have used meters to express length.
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And in the denominator, we have used kilometers.
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In order to perform our calculation, these should have the same units.
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So we must convert our radius from kilometers two meters before we continue.
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We can recall that one kilometer is equal to 1,000 meters, or equivalently we could write this as one times 10 to the power of three meters.
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So we can simply multiply our value of ๐ by 10 to the power of three to convert it from kilometers to meters, giving us a radius of 2,630 times 10 to the power of three meters.
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Now that weโve converted our units, we can see that we can express meters cubed per second squared as meters multiplied by meters multiplied by meters divided by seconds squared, and we can write meters squared as meters multiplied by meters.
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And we can now see that we can cancel two of these meters terms from the numerator and denominator of this fraction, leaving us with units of just meters per second squared, and these are the correct units for acceleration.
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We can now continue and substitute our values into this equation.
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This gives us ๐ is equal to 6.67 times 10 to the power of negative 11 meters cubed per kilogram-seconds squared multiplied by 1.48 times 10 to the power of 23 kilograms divided by 2,630 times 10 to the power of three meters squared.
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Evaluating this expression, we get ๐ is equal to 1.427 and so on meters per second squared.
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Our final thing to do is to give our answer to two decimal places.
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1.427 and so on to two decimal places is just equal to 1.43.
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So ๐ is equal to 1.43 meters per seconds squared.
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This is the acceleration that the astronaut experiences on the surface of Ganymede and is also known as Ganymedeโs surface gravity.
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So, the surface gravity on Ganymede to two decimal places is equal to 1.43 meters per seconds squared.