Question

on the low end of the scale, meteorites have a mass of at least 100 kg. the mass of earth is approximately 5.972·10²⁴ kg. calculate the force of gravity between earth and a meteorite at the edge of earth’s atmosphere given the distance from outer space to the center of the earth is 6.478·10⁶ m. remember that the gravitational constant, g, is 6.6743·10⁻¹¹ m³/kg·s². 949.8 n; 1.423·10⁹ n; 6.153·10⁹ n; 2.58·10¹⁷ n

Explanation:

Step1: Recall gravitational force formula

Newton's law of universal gravitation: $F = G\frac{m_1m_2}{r^2}$
Where:
$G=6.6743\times10^{-11}\ \text{m}^3/\text{kg}\cdot\text{s}^2$, $m_1=5.972\times10^{24}\ \text{kg}$, $m_2=100\ \text{kg}$, $r=6.478\times10^6\ \text{m}$

Step2: Substitute values into formula

$$ F = 6.6743\times10^{-11} \times \frac{5.972\times10^{24} \times 100}{(6.478\times10^6)^2} $$

Step3: Calculate denominator first

$(6.478\times10^6)^2 = 6.478^2\times10^{12} \approx 41.964\times10^{12} = 4.1964\times10^{13}$

Step4: Calculate numerator

$5.972\times10^{24} \times 100 = 5.972\times10^{26}$
$6.6743\times10^{-11} \times 5.972\times10^{26} \approx 39.86\times10^{15} = 3.986\times10^{16}$

Step5: Divide numerator by denominator

$F \approx \frac{3.986\times10^{16}}{4.1964\times10^{13}} \approx 0.9498\times10^{3} = 949.8\ \text{N}$

Answer:

949.8 N