Предмет: Физика, автор: oksanastrelcov

Гагарин на космическом корабле двигался по круговой
орбите,, среднее расстояние которой от поверхности земли =251 км.
определите период обращения корабля
вокруг земли.

Ответы

Автор ответа: toft1

GM/R^2 = g => GM = gR^2 чтобы не возиться с массой Земли.
Равенство гравитационного и центростремительного ускорений на круговой орбите:
gR^2/(R+h) = omega^2(R+h) = 4pi^2(R+h)/T^2 (поскольку угловая скорость
omega = 2pi/T)
позволяет определить T:
T = 2pi*sqrt((R+h)^3/(gR^2)).
Подставляя h = 0.251e-06 м
R = 6.371e06
g = 9.8
получаем:
T = 6.28/(6.371e06)*sqrt((6.371+0.251)^3*1e18/9.8) = 0.98e-06*sqrt(290/9.8)1e09 = 5280 сек = 5280/3600 = 1.47 часа

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