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PHYSICS, M5 2019 VCE 5*

Navigation in vehicles or on mobile phones uses a network of global positioning system (GPS) satellites. The GPS consists of 31 satellites that orbit Earth.

In December 2018, one satellite of mass 2270 kg, from the GPS Block \(\text{IIIA}\) series, was launched into a circular orbit at an altitude of \(20\ 000\) km above Earth's surface.

  1. Identify the type(s) of force(s) acting on the satellite and the direction(s) in which the force(s) must act to keep the satellite orbiting Earth.   (2 marks)
  1. Calculate the period of the satellite to three significant figures. You may use data from the table below in your calculations. Show your working.   (3 marks)
mass of satellite 2.27 × 10\(^3\) kg
altitude of satellite above Earth's surface 2.00 × 10\(^7\) m
Show Answers Only

a.    → Only force acting on a satellite is \(F_g\), the force due to gravity.

→ This force acts on the satellite directly towards the centre of the Earth.v

b.    \(4.25 \times 10^4\ \text{s}\)

Show Worked Solution

a.    → Only force acting on a satellite is \(F_g\), the force due to gravity.

→ This force acts on the satellite directly towards the centre of the Earth.
 

♦ Mean mark (a) 46%.
COMMENT: Thrust force is not a requirement for orbit. 

b.    \(\dfrac{r^3}{T^2}=\dfrac{GM}{4\pi^2}\ \ \Rightarrow \ \ T= \sqrt{\dfrac{4 \pi^2r^3}{GM}}\)

\(T=\sqrt{\dfrac{4\pi^2 \times (6.371 \times 10^6 + 2 \times 10^7)^3}{6.67 \times 10^{-11} \times 6 \times 10^{24}}}=42\ 533=4.25\times 10^4\ \text{s}\)

PHYSICS, M5 2020 VCE 4*

The Ionospheric Connection Explorer (ICON) space weather satellite, constructed to study Earth's ionosphere, was launched in October 2019. ICON will study the link between space weather and Earth's weather at its orbital altitude of 600 km above Earth's surface. Assume that ICON's orbit is a circular orbit. 

  1. Calculate the orbital radius of the ICON satellite.   (1 mark)

  2. Calculate the orbital period of the ICON satellite correct to three significant figures. Show your working.   (4 marks)

  3. Explain how the ICON satellite maintains a stable circular orbit without the use of propulsion engines.   (2 marks)

Show Answers Only

a.    \(r_{orbit}= 6.971 \times 10^6\ \text{m}\)

b.    \(5780\ \text{s}\)

c.    → The icon satellite is only subject to the gravitational force of the Earth.

→ This force is a centripetal force of constant magnitude and hence the satellite maintains a stable circular orbit. 

Show Worked Solution

a.    Radius of orbit is equal to the altitude plus the radius of the Earth.

\(r_{orbit}=600\ 000\ \text{m} + 6.371 \times 10^6\ \text{m} = 6.971 \times 10^6\ \text{m}\)

 

b.     \(\dfrac{r^3}{T^2}\) \(=\dfrac{GM}{4\pi^2}\)
  \(T\) \(=\sqrt{\dfrac{4\pi^2r^3}{GM}}\)
    \(=\sqrt{\dfrac{4\pi^2 \times (6.971 \times 10^6)^3}{6.67 \times 10^{-11} \times 6 \times 10^{24}}}\)
    \(=5780\ \text{s}\ \ \text{(3 sig.fig.)}\)

 

c.    → The icon satellite is only subject to the gravitational force of the Earth.

→ This force is a centripetal force of constant magnitude and hence the satellite maintains a stable circular orbit. 

♦♦♦ Mean mark (c) 23%.

PHYSICS, M5 2021 VCE 3

To calculate the mass of distant pulsars, physicists use Newton's law of universal gravitation and the equations of circular motion.

The planet Phobetor orbits pulsar PSR B1257+12 at an orbital radius of 6.9 × 10\(^{10}\) m and with a period of 8.47 × 10\(^6\) s.

Assuming that Phobetor follows a circular orbit, calculate the mass of the pulsar. Show all your working.  (3 marks)

Show Answers Only

\(M=2.71\ \times 10^{30}\ \text{kg}\)

Show Worked Solution
\(\dfrac{r^3}{T^2}\) \(=\dfrac{GM}{4\pi^2}\)  
\(M\) \(=\dfrac{4\pi^2 r^3}{G T^2}\)  
  \(=\dfrac{4 \pi^2 \times (6.9 \times 10^{10})^3}{6.67 \times 10^{-11} \times (8.47 \times 10^6)^2}\)  
  \(=2.71 \times 10^{30}\ \text{kg}\)  

PHYSICS, M5 2022 VCE 2

There are over 400 geostationary satellites above Earth in circular orbits. The period of orbit is one day (86 400 seconds). Each geostationary satellite remains stationary in relation to a fixed point on the equator. Figure 2 shows an example of a geostationary satellite that is in orbit relative to a fixed point, \(\text{X}\), on the equator.
 

  1. Explain why geostationary satellites must be vertically above the equator to remain stationary relative to Earth's surface.  (2 marks)
  1.  Using  \(G=6.67 \times 10^{-11}\ \text{N m}^{2}\ \text{kg}^{-2}, M_{\text{E}} = 5.98 \times 10^{24}\ \text{kg}\)  and  \(R_{\text{E}} = 6.37 \times 10^{6}\ \text{m}\), show that the altitude of a geostationary satellite must be equal to \(3.59 \times 10^{7}\ \text{m}\).  (4 marks)
  1. Calculate the speed of an orbiting geostationary satellite.  (3 marks)

Show Answers Only

a.   Reasons the satellite must orbit relative to a fixed point on the equator:

→ So that it is in the same rotational plane/axis of the Earth. 

→ The gravitational force on the satellite must be directly towards the centre of the Earth and it must orbit at a set altitude to the Earth where the speed of the satellite matches the rotational speed of the Earth.

→ As the speed and direction of the satellite matches that of the Earth, it will remain stationary relative to the motion of the Earth.

b.    See worked solutions

c.    3076 ms\(^{-1}\)

Show Worked Solution

a.   Reasons the satellite must orbit relative to a fixed point on the equator:

→ So that it is in the same rotational plane/axis of the Earth. 

→ The gravitational force on the satellite must be directly towards the centre of the Earth and it must orbit at a set altitude to the Earth where the speed of the satellite matches the rotational speed of the Earth.

→ As the speed and direction of the satellite matches that of the Earth, it will remain stationary relative to the motion of the Earth.
 

♦♦♦ Mean mark (a) 19%.
b.     \(\dfrac{r^3}{T^2}\) \(=\dfrac{GM}{4\pi^2}\)
  \(r\) \(=\sqrt[3]{\dfrac{GMT^2}{4\pi^2}}\)
    \(=\sqrt[3]{\dfrac{6.67 \times 10^{-11} \times 5.98 \times 10^{24} \times (86\ 400)^2}{4\pi^2}}\)
    \(=42.250 \times 10^6\ \text{m}\)

 
\(\therefore \ \text{Altitude}\ =42.250 \times 10^6 -6.37 \times 10^6 =3.59 \times 10^7\ \text{m}\)
 

c.     \(v\) \(=\sqrt{\dfrac{GM}{r}}\)
    \(=\sqrt{\dfrac{6.67 \times 10^{-11} \times 5.98 \times 10^{24}}{42.25 \times 10^6}}\)
    \(=3073\ \text{ms}^{-1}\)

PHYSICS, M5 2023 VCE 2

Phobos is a small moon in a circular orbit around Mars at an altitude of 6000 km above the surface of Mars.

The gravitational field strength of Mars at its surface is 3.72 N kg\(^{-1}\). The radius of Mars is 3390 km.

  1. Show that the gravitational field strength 6000 km above the surface of Mars is 0.48 N kg\(^{-1}\).   (2 marks)
  1. Calculate the orbital period of Phobos. Give your answer in seconds.  (3 marks)
  1. Phobos is very slowly getting closer to Mars as it orbits.
  2. Will the orbital period of Phobos become shorter, stay the same or become longer as it orbits closer to Mars? Explain your reasoning.  (2 marks)

Show Answers Only

a.    \(0.48\ \text{N kg}^{-1}\)

b.    \(2.77 \times 10^4\ \text{s}\)

c.    The orbital period will become shorter.

Show Worked Solution

a.    \(\text{Find the mass of Mars:}\)

         \(g\) \(=\dfrac{GM}{r^2}\)
  \(M\) \(=\dfrac{gr^2}{G}\)
    \(=\dfrac{3.72 \times (3390 \times 10^3)^2}{6.67 \times 10^{-11}}\)
    \(=6.409 \times 10^{23}\)

 
\(\text{Find the gravitational field strength where}\ r= 9390\ \text{km:}\) 

         \(g\) \(=\dfrac{GM}{r^2}\)
    \(=\dfrac{6.67 \times 10^{-11} \times 6.41 \times 10^{23}}{(9390000)^2}\)
    \(=0.48\ \text{N kg}^{-1}\)

 

b.     \(\dfrac{r^3}{T^2}\) \(=\dfrac{GM}{4\pi^2}\)
  \(T\) \(=\sqrt{\dfrac{4\pi^2 r^3}{GM}}\)
    \(=\sqrt{\dfrac{4\pi^2 (9390000)^3}{6.67 \times 10^{-11} \times 6.409 \times 10^{23}}}\)
    \(=27652\ \text{s}\)
    \(=2.77 \times 10^4\ \text{s}\)

♦ Mean mark (b) 53%. 
COMMENT: Rounded calculations, such as using 6.41 × 10^23 were marked as wrong.

c.    \(T=\sqrt{\dfrac{4\pi^2 r^3}{GM}}\)

\(T \propto \sqrt{r^3}\)

→ \(\text{decreasing the orbital radius will also decrease the orbital period.}\)

PHYSICS, M5 2023 HSC 5 MC

An exoplanet is in an elliptical orbit, moving in the direction shown. The distances between consecutive positions \(P, Q, R\) and \(S\) are equal.
 


 

Between which two points is the exoplanet's travel time the greatest?

  1. \(P\) and \(Q\)
  2. \(Q\) and \(R\)
  3. \(R\) and \(S\)
  4. \(S\) and \(P\)
Show Answers Only

\(D\)

Show Worked Solution

→ As the exoplanet is in an elliptical orbit it will travel the slowest when it is the furthest away from the star.

→ As all distances between the points are equal, the exoplanet will have the greatest travel time between \(S\) and \(P\) where it moves the slowest.

\(\Rightarrow D\)

PHYSICS, M5 EQ-Bank 29

A student used the following scale diagram to investigate orbital properties. The diagram shows a planet and two of its moons, `V` and `W`. The distances between each of the moons and the planet are to scale while the sizes of the objects are not.
 

Complete the table to compare the orbital properties of Moon `V` and Moon `W`. Show relevant calculations in the space below the table.  (4 marks)
 

Show Answers Only

\begin{array} {|l|c|c|c|}
\hline   & \textit{Orbital radius} & \textit{Orbital period} & \textit{Orbital velocity} \\
\hline \text{Quantitative Comp.} & 3.0 & 5.2 & 0.58 \\
\hline \text{Qualitative Comp.} & \text{Larger} & \text{Larger} & \text{Slower} \\
\hline \end{array}

 
→ Comparing radii:

`(r_(w))/(r_(v))=(10)/(3.3)=3`
 

→ As moons `V` and `W` are both orbiting the same body, the ratio `(r^3)/(T^2)`will be the same for both.

→ Comparing orbital periods:

`(r_(w)^(3))/(T_(w)^(2))` `=(r_(v)^(3))/(T_(v)^(2))`  
`((T_(w))/(T_(v)))^(2)` `=((r_(w))/(r_(v)))^(3)=3^3`  
`(T_(w))/(T_(v))` `=sqrt(27)=5.2`  

 
→ Comparing orbital velocities.

`v=(2pir)/(T)`

 
→ Using the calculations for radius and period:

`v_(w)` `=(2pir_(w))/(T_(w))`  
  `=(2pi(3.0xxr_(v)))/(5.2xxT_(v))`  
  `=(3.0)/(5.2)xx(2pir_(v))/(T_(v))`  
  `=(3.0)/(5.2)xxv_(v)`  
`(v_(w))/(v_(v))` `=0.58`  
Show Worked Solution

\begin{array} {|l|c|c|c|}
\hline   & \textit{Orbital radius} & \textit{Orbital period} & \textit{Orbital velocity} \\
\hline \text{Quantitative Comp.} & 3.0 & 5.2 & 0.58 \\
\hline \text{Qualitative Comp.} & \text{Larger} & \text{Larger} & \text{Slower} \\
\hline \end{array}

 
→ Comparing radii:

`(r_(w))/(r_(v))=(10)/(3.3)=3`
 

→ As moons `V` and `W` are both orbiting the same body, the ratio `(r^3)/(T^2)`will be the same for both.

→ Comparing orbital periods:

`(r_(w)^(3))/(T_(w)^(2))` `=(r_(v)^(3))/(T_(v)^(2))`  
`((T_(w))/(T_(v)))^(2)` `=((r_(w))/(r_(v)))^(3)=3^3`  
`(T_(w))/(T_(v))` `=sqrt(27)=5.2`  

 
→ Comparing orbital velocities.

`v=(2pir)/(T)`

 
→ Using the calculations for radius and period:

`v_(w)` `=(2pir_(w))/(T_(w))`  
  `=(2pi(3.0xxr_(v)))/(5.2xxT_(v))`  
  `=(3.0)/(5.2)xx(2pir_(v))/(T_(v))`  
  `=(3.0)/(5.2)xxv_(v)`  
`(v_(w))/(v_(v))` `=0.58`  

PHYSICS, M5 EQ-Bank 21

Comet Hale-Bopp is a long period comet that is believed to come from the Oort cloud which lies far beyond the outermost planets. In our solar system, Hale-Bopp travels in an elliptical orbit around the Sun and spends most of its time beyond the outermost planets.

Using a diagram, explain how the motion of Comet Hale-Bopp in its orbit supports Kepler's second law.  (3 marks)

Show Answers Only

→ Kepler’s Second Law states that a line between the comet and the sun will sweep an equal area in equal time.

→ When the comet is closer to the sun, it must travel with a greater velocity in its orbit compared to when it is far away to sweep the same area in the same time (i.e. such that `A_1=A_2`.

→ Comet Hale-Bopp supports this as it spends most of its time beyond the outermost planets. It therefore travels at a slower speed compared to when it is close to the sun. This reduced velocity allows it to sweep the same area compared to when it is closer to the sun. 

Show Worked Solution

→ Kepler’s Second Law states that a line between the comet and the sun will sweep an equal area in equal time.

→ When the comet is closer to the sun, it must travel with a greater velocity in its orbit compared to when it is far away to sweep the same area in the same time (i.e. such that `A_1=A_2`.

→ Comet Hale-Bopp supports this as it spends most of its time beyond the outermost planets. It therefore travels at a slower speed compared to when it is close to the sun. This reduced velocity allows it to sweep the same area compared to when it is closer to the sun. 

PHYSICS, M5 EQ-Bank 15 MC

The table shows data about the solar system.

\begin{array} {|l|c|c|}
\hline  \textit{Planet} & \textit{Average distance from the sun (AU)} & \textit{Period (days)} \\
\hline \text{Venus} & 0.72 & 224.7 \\
\hline \text{Earth} & 1 & 365 \\
\hline \text{Mars} & 1.54 & 686.2 \\
\hline \end{array}

What would be the period of another planet if it orbited the Sun at an average distance of 4.5 AU ?

  1. `7.9 xx 10^(2)` days
  2. `1.5 xx 10^(3)` days
  3. `1.6 xx 10^(3)` days
  4. `6.6 xx 10^(3)` days
Show Answers Only

`B`

Show Worked Solution

→ `(r^3)/(T^2)=(GM)/(4pi^2)`

→ Since all planets are orbiting the same central body with mass, `M`, `G` and `4pi^2` are constants.

→ The ratio `(r^3)/(T^2)` will be the same for all planets in the solar system.

`(r^3)/(T^2)` `=(r_e^3)/(T_e^2)`  
`(4.5^3)/(T^2)` `=(1^3)/(365^2)`  
`T^2` `=(365^2)/(4.5^3)`  
`:.T` `=sqrt((365^2)/(4.5^3))`  
  `=1462\ text{days}`  
  `~~1.5xx10^3\ text{days}`  

 
`=>B`

PHYSICS, M5 2016 HSC 14 MC

A satellite orbits Earth with period `T`. An identical satellite orbits the planet Xerus which has a mass four times that of Earth. Both satellites have the same orbital radius `r`.
 

What is the period of the satellite orbiting Xerus?

  1. `(T)/4`
  2. `(T)/2`
  3. `2T`
  4. `4T`
Show Answers Only

`B`

Show Worked Solution

Applying Kepler’s Third Law:

`(r^(3))/(T^(2))` `=(GM)/(4pi^(2))`  
`r^3` `=(GMT^(2))/(4pi^(2))`  

 
Since satellites have the same orbital radius:

`(G(M_(E))(T_E)^(2))/(4pi^(2))` `=(G(M_(X))(T_(X))^(2))/(4pi^(2))`  
`(M_(E))(T_E)^(2)` `=(M_(X))(T_(X))^(2)`  
`((T_(X))^(2))/((T_E)^(2))` `=(M_(E))/(M_(X))`  
  `=1/4\ \ (M_X=4xxM_E\ text{(given)})`  
`(T_(X))/(T_E)` `=(1)/(2)`  
`T_X` `=(T_E)/2`  

 
`=>B`


♦ Mean mark 45%.

PHYSICS, M5 2019 HSC 11 MC

A dwarf planet orbits the sun with a period of 40 000 years.

The average distance from the sun to Earth is one astronomical unit.

What is the average distance between this dwarf planet and the sun in astronomical units?

  1. `34`
  2. `200`
  3. `1170`
  4. `8 × 10^(6)`
Show Answers Only

`C`

Show Worked Solution

Kepler’s Third Law:

`(r^3)/(T^2)`  is constant for objects orbiting the same central body. 

`(r_text{Planet}^3)/(T_text{Planet}^2)` `=(r_text{Earth}^3)/(T_text{Earth}^2)`  
`(r_text{Planet}^3)/(40\ 000^2)` `= 1`  
`r_text{Planet}` `=root(3)(40\ 000^2)`  
  `=1170`  

 
`=>C`

PHYSICS, M5 2021 HSC 25

A satellite is launched from the surface of Mars into an orbit that keeps it directly above a position on the surface of Mars.

Mass of Mars = `6.39 xx 10^(23)` kg

Length of Martian day = 24 hours and 40 minutes

  1. Identify TWO energy changes as the satellite moves from the surface of Mars into orbit.   (2 marks)
  1. Calculate the orbital radius of the satellite.   (3 marks)
Show Answers Only

a.   The kinetic energy and gravitational potential energy of the satellite both increase as it moves into orbit.

b.   `2.0 xx10^(7)` m

Show Worked Solution

a.   The kinetic energy and gravitational potential energy of the satellite both increase as it moves into orbit.

b.   `(r^(3))/(T^(2))` `=(GM)/(4pi^(2))`  
`r^(3)` `=((6.67 xx10^(-11)xx6.39 xx10^(23))/(4pi^(2)))((24+(40)/(60))xx60 xx60)^(2)`  
`:. r` `=2.0 xx10^(7)` m