The International Space Station, with a mass of 4.19x10^5 kg, is in uniform circular motion with a radius of 6.8x10^6m as measured from the center of the earth. Calculate the speed of the ISS as it revolves around earth

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A body in revolution around the Earth is kept in its orbit due to a centripetal force that acts on it accelerating it towards the Earth. The centripetal force is equivalent to the gravitational force of attraction between the body and the Earth. If the speed of the object is v and its radius is R, the centripetal force required is `m*v^2/R` . The gravitational force of attraction between the object and the Earth is `(m*Me*G)/R^2` . Equating the two gives `v^2 = (G*Me)/R` or `v = sqrt((G*Me)/R)`

The mass of the Earth is `5.97219*10^24` kg and the radius of the International Space Station's orbit is `6.8x10^6` m. The gravitational constant G = `6.673*10^-11` m^3*kg^-1*s^-2. Using these gives the speed of the International Space Station as `sqrt((6.673*10^-11*5.97219*10^24)/(6.8*10^6))` = 7655.48 m/s^2

The International Space Station is moving around the Earth at 7655.48 m/s^2

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