Question

determine the force of gravitational attractions between the Earth 5.98 x 10^24 kg and a 70kg boy is standing at sea level, a distance og 6.38 x 10^6m Earth's center ( the universal gravitational constant (G) = 6.6726 x 10 ( raise to power) 11 N/m/kg^ 2)​

Answer 1

Answer:

6.862035456*10^26 m/s^2

Step-by-step explanation:

We know that:

m=70kg

M=Mass of earth=5.98*10^24kg

R=Distance between object's and earth's center=6.38*10^6m

G=6.6726*10^-11 N/m/kg^2

Thus, we can use the equation $F_g=\frac{GmM}{R^2}$:

$F_g=\frac{(6.6726*10^-11)(70)(5.98*10^{24})}{(6.38*10^{-6})^2}\\\\F_g=6.862035456*10^{26}$

This means the force of gravitational attractions is 6.862035456*10^26 m/s^2