The International Space Station (ISS) orbits the Earth at an altitude of around 410 km. The ISSdoes not fall towards the Earth, and the astronauts & objects are floating weightlessly. On Earth’ssurface, objects fall with a gravitational acceleration of approximately 9.81 m/s2.(a) Show that the following equation gives the percentage P of how much the grav. accelerationdecreases between Earth’s surface (R: Earth’s radius, 6371 km) and an object at altitude z:P (z) = 1 −( 11 + zR)2(b) What is P (z) for the altitude of the ISS, and why are objects weightless nevertheless?(c) At what distance does the grav. acceleration become only 1% of that on Earth’s surface?
Question
The International Space Station (ISS) orbits the Earth at an altitude of around 410 km. The ISSdoes not fall towards the Earth, and the astronauts & objects are floating weightlessly. On Earth’ssurface, objects fall with a gravitational acceleration of approximately 9.81 m/s2.(a) Show that the following equation gives the percentage P of how much the grav. accelerationdecreases between Earth’s surface (R: Earth’s radius, 6371 km) and an object at altitude z:P (z) = 1 −( 11 + zR)2(b) What is P (z) for the altitude of the ISS, and why are objects weightless nevertheless?(c) At what distance does the grav. acceleration become only 1% of that on Earth’s surface?
Solution
(a) The gravitational acceleration at a distance r from the center of the Earth is given by g(r) = GM/r^2, where G is the gravitational constant and M is the mass of the Earth. At the surface of the Earth, r = R and g(R) = GM/R^2 = g0, where g0 is the gravitational acceleration at the surface of the Earth. At an altitude z above the surface of the Earth, r = R + z and g(R + z) = GM/(R + z)^2.
The percentage P of how much the gravitational acceleration decreases between the Earth's surface and an object at altitude z is given by P(z) = 1 - g(R + z)/g0. Substituting the expressions for g(R + z) and g0 into this equation gives P(z) = 1 - (GM/(R + z)^2) / (GM/R^2) = 1 - (R^2/(R + z)^2) = 1 - (1 + z/R)^-2.
(b) Substituting z = 410 km = 0.410 Mm into the equation for P(z) gives P(0.410) = 1 - (1 + 0.410/6371)^-2 ≈ 0.059 or 5.9%. This means that the gravitational acceleration at the altitude of the ISS is about 5.9% less than at the surface of the Earth. Objects are weightless in the ISS because they are in free fall. The ISS is constantly falling towards the Earth, but it also has a horizontal velocity that causes it to continually miss the Earth. This is what it means to be in orbit.
(c) Setting P(z) = 0.01 and solving for z gives z = R(1/(1 - 0.01)^0.5 - 1) ≈ 0.005R or 31.9 Mm. This means that the gravitational acceleration becomes only 1% of that on the Earth's surface at a distance of about 31.9 Mm from the surface of the Earth.
Similar Questions
The International Space Station (ISS) orbits the Earth at an altitude of around 410 km. The ISSdoes not fall towards the Earth, and the astronauts & objects are floating weightlessly. On Earth’ssurface, objects fall with a gravitational acceleration of approximately 9.81 m/s2.(a) Show that the following equation gives the percentage P of how much the grav. accelerationdecreases between Earth’s surface (R: Earth’s radius, 6371 km) and an object at altitude z:P (z) = 1 −( 11 + zR)2(b) What is P (z) for the altitude of the ISS, and why are objects weightless nevertheless?(c) At what distance does the grav. acceleration become only 1% of that on Earth’s surface?
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