Question 3:
Figure 2 shows the orbit of planet Uranus.
Describe the change in linear speed of planet Uranus when it moves from point A to point B.
Answer:
Keplers’ Second Law states that a line connecting a planet to the Sun sweep out equal areas in equal times.
The shorter the distance of a planet from the Sun, the faster the velocity of the planet. From A to B, the speed of planet Uranus increases to a maximum value and then decreases again.
Figure 2 shows the orbit of planet Uranus.
Describe the change in linear speed of planet Uranus when it moves from point A to point B.
Answer:
Keplers’ Second Law states that a line connecting a planet to the Sun sweep out equal areas in equal times.
The shorter the distance of a planet from the Sun, the faster the velocity of the planet. From A to B, the speed of planet Uranus increases to a maximum value and then decreases again.
Question 4:
Figure 3 shows the Earth, the Moon and a satellite.
(a) Which pair of bodies experience the smallest gravitational force?
Give a reason for your answer.
(b) Calculate the gravitational force between the Earth and the satellite.
[Mass of the Earth = 5.97 × 1024 kg, mass of satellite = 1.2 × 103 kg, distance between centre of the Earth and centre of the satellite = 7.87 × 106 m]
Answer:
(a)
Satellite and the Moon as a pair.
The result from the multiplication of the mass of the pair is the smallest. Distance separating the pair is the largest.
(b)
$$ \begin{aligned} F & =\frac{G M m}{r^2} \\ & =\frac{\left(6.67 \times 10^{-11}\right) \times\left(5.97 \times 10^{24}\right) \times\left(1.2 \times 10^3\right)}{\left(7.87 \times 10^6\right)^2} \\ & =7.71 \times 10^3 \mathrm{~N} \end{aligned} $$
Figure 3 shows the Earth, the Moon and a satellite.
(a) Which pair of bodies experience the smallest gravitational force?
Give a reason for your answer.
(b) Calculate the gravitational force between the Earth and the satellite.
[Mass of the Earth = 5.97 × 1024 kg, mass of satellite = 1.2 × 103 kg, distance between centre of the Earth and centre of the satellite = 7.87 × 106 m]
Answer:
(a)
Satellite and the Moon as a pair.
The result from the multiplication of the mass of the pair is the smallest. Distance separating the pair is the largest.
(b)
$$ \begin{aligned} F & =\frac{G M m}{r^2} \\ & =\frac{\left(6.67 \times 10^{-11}\right) \times\left(5.97 \times 10^{24}\right) \times\left(1.2 \times 10^3\right)}{\left(7.87 \times 10^6\right)^2} \\ & =7.71 \times 10^3 \mathrm{~N} \end{aligned} $$