A space probe of mass 5-60 × 10^(-3) kg is in orbit at a height of 3.70 × 10^(-6) m above the surface of Mars - Scottish Highers Physics - Question 3 - 2015

To calculate the gravitational force between the space probe and Mars, we can use the formula:

$F = \frac{GMm}{r^2}$

Where:

• $G = 6.67 \times 10^{-11} \text{ N m}^2/\text{kg}^2$ (gravitational constant)
• $M = 6.42 \times 10^{23} \text{ kg}$ (mass of Mars)
• $m = 5.60 \times 10^{-3} \text{ kg}$ (mass of the probe)
• $r = R + h$ (distance from the center of Mars to the probe)

Calculating $r$:

• Radius of Mars, $R = 3.39 \times 10^{6} \text{ m}$
• Height of the probe above the surface, $h = 3.70 \times 10^{-6} \text{ m}$

Thus: $r = 3.39 \times 10^{6} + 3.70 \times 10^{-6} \approx 3.39 \times 10^{6} \text{ m}$

Now, substitute these values into the gravitational force formula:

$F = \frac{(6.67 \times 10^{-11})(6.42 \times 10^{23})(5.60 \times 10^{-3})}{(3.39 \times 10^{6})^2}$

Calculating this gives: $F \approx 4.77 \times 10^{1} \text{ N}$