According to Newton's law of universal gravitation, "F = G \\frac{m_1 m_2}{r^2}", where "G = 6.67 \\cdot 10^{-11} m^3 kg^{-1}s^{-2}" is the gravitational constant, "m_1, m_2" - masses of the two bodies and "r" is the distance between them.
Let "R_e" be the radius of Earth and "h = 6 \\cdot 10^6 m" the height above the surface of the Earth. Then, the force acting on 1 kg mass is "F = G \\frac{M_e}{(R_e+h)^2}", where "M_e = 5.97 \\cdot 10^{24} kg" is the mass of the Earth. Calculating last expression, obtain "F = 2.6 N". This quantity is actually the acceleration due to Earth's gravity at given distance. Since it is quite far from the surface of the Earth, is it much lower than the gravity of the Earth "g = 9.81 \\frac{m}{s^2}".
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