Question Video: Finding the Local Acceleration Due to Gravity at a Position Above Earth’s Surface | Nagwa Question Video: Finding the Local Acceleration Due to Gravity at a Position Above Earth’s Surface | Nagwa

Question Video: Finding the Local Acceleration Due to Gravity at a Position Above Earth’s Surface Physics • First Year of Secondary School

The International Space Station orbits Earth at a distance of 409 km above the surface. Earth has a mass of 5.97 × 10²⁴ kg and a radius of 6,370 km. What is the local acceleration due to gravity at the height at which the International Space Station orbits? Give your answer to two decimal places.

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Video Transcript

The International Space Station orbits Earth at a distance of 409 kilometers above the surface. Earth has a mass of 5.97 times 10 to the 24 kilograms and a radius of 6,370 kilometers. What is the local acceleration due to gravity at the height at which the International Space Station orbits? Give your answer to two decimal places.

So here we have the International Space Station in orbit around Earth. And we’re asked to find the local acceleration due to gravity. We know that the International Space Station experiences an attractive force towards Earth because of the force of gravity and that this depends on Earth’s mass, which is given to us as 5.97 times 10 to the 24 kilograms. We’re also told that the International Space Station orbits Earth at a distance 409 kilometers above the surface and also that Earth has a radius of 6,370 kilometers.

So we need to recall an equation that relates local acceleration due to gravity to mass and distance. That equation is 𝑎 equals 𝐺𝑀 over 𝑟 squared, where 𝑎 is the local acceleration due to gravity, which is the quantity we’re trying to find. 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 the mass of Earth, which is 5.97 times 10 to the 24 kilograms. And 𝑟 is the distance to Earth’s center of mass.

To get the distance to Earth’s center of mass, we need to add the height above Earth’s surface, which is 409 kilometers, to the distance between Earth’s surface and the center of Earth, which is the same as the radius, or 6,370 kilometers. Therefore, 𝑟 is equal to 6,370 kilometers plus 409 kilometers, which is equal to 6,779 kilometers. We need this to be in meters. So recall that one kilometer is equal to 1,000 meters. And therefore, 𝑟 is equal to 6,779,000 meters.

We can now put all of this together, so we have 𝑎 is equal to 𝐺𝑀 over 𝑟 squared. So we have that 𝑎 is equal to 6.67 times 10 to the minus 11 meters cubed per kilogram second squared times 5.97 times 10 to the 24 kilograms divided by 6,779,000 meters squared, which comes out to 8.665 meters per second squared. Now, we’re asked to give this to two decimal places, so that comes to 8.67 meters per second squared.

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