Gravitational field — practice questions

Identify known values: $m_1 = 70 \text{ kg}$, $m_2 = 1200 \text{ kg}$, $r = 3.0 \text{ m}$, $G = 6.67 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}$.

Recall Newton's Law of Gravitation formula: $F = G \frac{m_1 m_2}{r^2}$.

Substitute the values into the formula: $F = (6.67 \times 10^{-11}) \times \frac{(70)(1200)}{(3.0)^2}$

Calculate the product of masses and square of distance: $F = (6.67 \times 10^{-11}) \times \frac{84000}{9.0}$

Perform the final calculation: $F = (6.67 \times 10^{-11}) \times 9333.33 \dots$ $F \approx 6.22 \times 10^{-7} \text{ N}$

The gravitational force between the student and the car is approximately $6.22 \times 10^{-7} \text{ N}$.

State the formula for gravitational field strength for a radial field: $g = \frac{GM}{r^2}$.

Identify the given values: $M = 6.42 \times 10^{23} \text{ kg}$ $r = 3390 \text{ km}$ $G = 6.67 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}$

Convert the radius from kilometres to metres, as the unit of G uses metres: $r = 3390 \text{ km} = 3390 \times 1000 \text{ m} = 3.39 \times 10^6 \text{ m}$.

Substitute the values into the formula: $g = \frac{(6.67 \times 10^{-11}) \times (6.42 \times 10^{23})}{(3.39 \times 10^6)^2}$

Calculate the numerator and the denominator: Numerator: $(6.67 \times 10^{-11}) \times (6.42 \times 10^{23}) = 4.28214 \times 10^{13}$ Denominator: $(3.39 \times 10^6)^2 = 11.4921 \times 10^{12}$

Perform the final division: $g = \frac{4.28214 \times 10^{13}}{11.4921 \times 10^{12}} \approx 3.726 \text{ N kg}^{-1}$

The gravitational field strength on the surface of Mars is approximately $3.73 \text{ N kg}^{-1}$ (or $3.73 \text{ m s}^{-2}$).