The diagram below shows a spaceship, with a mass of 3 500 kg, travelling in the vacuum of space in an orbit around the earth, with a mass of 5,98 × 10^24 kg at a constant speed - NSC Physical Sciences - Question 8 - 2016 - Paper 1

To calculate the gravitational force, we use the formula:

$F = G \frac{m_1 m_2}{r^2}$

Where:

• $G = 6.67 \times 10^{-11} \mathrm{N\cdot m^2\cdot kg^{-2}}$ (gravitational constant)
• $m_1 = 5.98 \times 10^{24} \mathrm{kg}$ (mass of the Earth)
• $m_2 = 3 500 \mathrm{kg}$ (mass of the spaceship)
• $r = 8.53 \times 10^6 \mathrm{m}$ (distance between the centers)

Substituting values:

$F = 6.67 \times 10^{-11} \frac{(5.98 \times 10^{24})(3500)}{(8.53 \times 10^6)^2} \approx 19 186.55 \mathrm{N}$

The force exerted by the spaceship on the earth is equal to the gravitational force calculated in QUESTION 8.2. This is because, according to Newton's third law of motion, every action has an equal and opposite reaction. Thus, the answer is: Equal to.

The law used is Newton’s third law of motion, which states that for every action, there is an equal and opposite reaction.

If the gravitational force increases by a factor of 4, we can express this using the gravitational force formula.

Since: $F' = 4F$

the new force will be:

$F' = G \frac{m_1 m_2}{(r')^2}$

Setting up the ratio:

$4F = G \frac{m_1 m_2}{(r')^2}$

We know: $F = G \frac{m_1 m_2}{r^2}$

Thus:

$4 \frac{m_1m_2}{r^2} = \frac{m_1 m_2}{(r')^2}$

This implies:

$(r')^2 = \frac{r^2}{4} \Rightarrow r' = \frac{r}{2}$

Substituting for $r = 8.53 \times 10^6 \mathrm{m}$:

$r' = \frac{8.53 \times 10^6}{2} = 4.265 \times 10^6 \mathrm{m}$