Question

Find the magnitude of the gravitational force
a 68.1 kg person would experience while
standing on the surface of Earth with a
mass of 5.98 × 1024 kg and a radius of
6.37 × 106 m. The universal gravitational
constant is 6.673 × 10−11 N · m2/kg2.
Answer in units of N

Answer 1

Find the magnitude of the gravitational force

a 68.1 kg person would experience while

standing on the surface of Earth with a

mass of $5.98 \times 10^{24} \, \mathrm{kg}$ and a radius of

$6.37 \times 10^{6} \, \mathrm{m}$. The universal gravitational

constant is $6.673 \times 10^{-11} \, \mathrm{N} \cdot \mathrm{m}^2/\mathrm{kg}^2$.

Answer in units of N

$$F = G \frac{m_1 m_2}{r^2} - \text{formula for the gravitation force}$$

where

$$m_1 = 68.1 \, \mathrm{kg}$$ $$m_2 = 5.98 \times 10^{24} \, \mathrm{kg}$$ $$r = 6.37 \times 10^6 \, \mathrm{m}$$ $$G = 6.673 \times 10^{-11} \quad (N \times \frac{m^2}{kg^2})$$ $$F = 6.673 \times 10^{-11} \frac{68.1 \times 5.98 \times 10^{24}}{(6.37 \times 10^6)^2} = 669.71 \, (N)$$