3. (5 pts) The International Space Station (ISS) orbits in a circular orbit 370

Question

3. (5 pts) The International Space Station (ISS) orbits in a circular orbit 370 km above the surface of the Earth. Using the radius of the Earth above and the altitude of the ISS, what is the acceleration due to gravity at this altitude? 4. (5 pts) Using the mass and radius of the Moon given on the first page, what is the escape speed from the surface of the Moon? That Is, how fast does an object have to be propelled upward from the surface of the Moon initially to get infinitely far away with zero speed? 5. (5 pts) Using the masses of the electron and proton given above, calculate the magnitude of the force of gravity between these two particles when they are 5.29 x 10^-11 m apart. This is the distance between the electron and proton in the ground state of the hydrogen atom. Expect an extremely small number.

Explanation / Answer

1)

a = GM/r^2 = 6.67*10^-11*6*10^24/((370+6371)*10^3)^2 = 8.81 m/s2 <---------answer

2)

for escape speed,

0.5*mv^2 = GMm/R

where M = mass of moon = 7.36*10^22 kg

R = 1.74*10^6 m

So, 0.5*v^2 = GM/R

So, v = sqrt(2GM/R)

So, v= sqrt(2*6.67*10^-11*7.36*10^22/(1.74*10^6)) = 2.38*10^3 m/s <---------answer

3)

F = GmM/R^2 = 6.67*10^-11*(9.1*10^-31)*(1.67*10^-27)/(5.29*10^-11)^2 = 3.62*10^-47 N <----------answer

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