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PHYSICS, M5 2016 HSC 21a

Why does orbital decay occur more rapidly for satellites in a low-Earth orbit than for satellites in other orbits?   (2 marks)

Show Answers Only

Satellites in a low-Earth orbit:

→ Encounter more atmospheric drag than satellites in higher orbits due to their lower altitude.

→ This reduces their orbital velocity which decreases orbital radius which in turn causes orbital decay.

Show Worked Solution

Satellites in a low-Earth orbit:

→ Encounter more atmospheric drag than satellites in higher orbits due to their lower altitude.

→ This reduces their orbital velocity which decreases orbital radius which in turn causes orbital decay.


Mean mark 59%.

PHYSICS, M5 2024 HSC 25

The mathematical model below shows the relationship between the orbital radius of a satellite and its period.

\(\dfrac{r^3}{T^2}=\dfrac{G M}{4 \pi^2}\)

  1. By considering gravitational force, show how this model can be derived.   (2 marks)

  2. A planet with five moons is discovered. The following graph is produced from observations of the orbital radius of the moons and their orbital periods, measured in Earth days.   
     

  1. Use the graph to calculate the mass of the planet.  (4 marks)

Show Answers Only

a.    See worked solutions

b.    \(9.9 \times 10^{25}\ \text{kg}\)

Show Worked Solution

a.    Centripetal force = gravitational force

\(F_c\) \(=F_g\)  
\(\dfrac{mv^2}{r}\) \(=\dfrac{GMm}{r^2}\)  
\(v^2\) \(=\dfrac{GM}{r}\ \ \text{where}\ \ v=\dfrac{2 \pi r}{T}\)  
\( \left(\dfrac{2\pi r}{T}\right)^2\) \(=\dfrac{GM}{r}\)  
\(\dfrac{4 \pi^2 r^2}{T^2}\) \(=\dfrac{GM}{r}\)  
\(\dfrac{r^3}{T^2}\) \(=\dfrac{GM}{4 \pi^2}\)  
  

b.    Find gradient of line through \((200, 0.16)\) and \((300, 0.24)\):

\(m=\dfrac{y_2-y_1}{x_2-x_1} = \dfrac{0.24-0.16}{300 \times 10^{12}-200 \times 10^{12}} = 8 \times 10^{-16}\ \text{days}^{2}\text{\km}^{3}\)

\(\Rightarrow\ \) To use \(\dfrac{r^3}{T^2} = \dfrac{GM}{4 \pi^2}\), \(r\) and \(T\) must be in SI units: 

\(8 \times 10^{-16} \times \dfrac{(24 \times 60 \times 60)^2}{1000^3} = 5.972 \times 10^{-15}\ \text{s}^{2}\text{\m}^{3}\)

\(\Rightarrow \ \dfrac{T^2}{r^3}= 5.972 \times 10^{-15}\ \text{s}^{2}\text{\m}^{3}\ \ \Rightarrow\ \ \dfrac{r^3}{T^2} = 1.6745 \times 10^{14}\)

\(\dfrac{r^3}{T^2}\) \(=\dfrac{GM}{4 \pi^2}\)  
\(M\) \(=\dfrac{r^3}{T^2} \times \dfrac{4 \pi^2}{G}\)  
  \(=1.6745 \times 10^{14} \times \dfrac{4 \pi^2}{6.67 \times 10^{-11}}\)  
  \(=9.9 \times 10^{25}\ \text{kg}\)  

PHYSICS, M5 2024 HSC 11 MC

A satellite is in a circular orbit.

What is the relationship between its orbital velocity, \(v\), and its orbital radius, \(r\)?

  1. \(v\) is directly proportional to the square of \(r\).
  2. \(v\) is inversely proportional to the square of \(r\).
  3. \(v\) is directly proportional to the square root of \(r\).
  4. \(v\) is inversely proportional to the square root of \(r\).
Show Answers Only

\(D\)

Show Worked Solution

→ The orbital velocity of a satellite is given by the equation: \(v=\sqrt{\dfrac{GM}{r}}\)

→ Hence  \(v \propto \dfrac{1}{\sqrt{r}}\)

→ The orbital velocity is inversely proportional to the square root of the orbital radius.

\(\Rightarrow D\)

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)

\begin{array}{|l|l|}
\hline \rule{0pt}{2.5ex}\text{mass of satellite} \rule[-1ex]{0pt}{0pt}& 2.27 \times 10^3 \ \text{kg} \\
\hline \rule{0pt}{2.5ex}\text{altitude of satellite above Earth's surface} \rule[-1ex]{0pt}{0pt}& 2.00 \times 10^7 \ \text{m} \\
\hline
\end{array}

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

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

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

         \(g\) \(=\dfrac{GM}{r^2}\)
    \(=\dfrac{6.67 \times 10^{-11} \times 6.41 \times 10^{23}}{(9\ 390\ 000)^2}\)
    \(=0.48\ \text{N kg}^{-1}\)

 

b.     \(\dfrac{r^3}{T^2}\) \(=\dfrac{GM}{4\pi^2}\)
  \(T\) \(=\sqrt{\dfrac{4\pi^2 r^3}{GM}}\)
    \(=\sqrt{\dfrac{4\pi^2 (9\ 390\ 000)^3}{6.67 \times 10^{-11} \times 6.409 \times 10^{23}}}\)
    \(=27\ 652\ \text{s}\)
    \(=2.77 \times 10^4\ \text{s}\)

♦ 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)

\begin{array}{|l|c|c|c|}
\hline
\rule{0pt}{2.5ex} & \textit{Orbital radius} & \textit{Orbital period} & \textit{Orbital velocity}\\
& \text{(\(W\) relative to \(V\))} \rule[-1ex]{0pt}{0pt}&  \text{(\(W\) relative to \(V\))}  &  \text{(\(W\) relative to \(V\))} \\
\hline
\rule{0pt}{3.5ex}\text{Quantitative}&\text{}&\text{}&\text{}\\
\text{comparison}\rule[-2ex]{0pt}{0pt}&\text{}&\text{}&\text{}\\
\hline
\rule{0pt}{3.5ex}\text{Qualitative}&\text{}&\text{}&\text{}\\
\text{comparison}\rule[-2ex]{0pt}{0pt}&\text{}&\text{}&\text{}\\
\hline
\end{array}

Show Answers Only

\begin{array} {|l|c|c|c|}
\hline  \rule{0pt}{2.5ex}  & \textit{Orbital radius} & \textit{Orbital period} & \textit{Orbital velocity} \\
\hline \rule{0pt}{2.5ex}\text{Quantitative Comp.} \rule[-1ex]{0pt}{0pt}& 3.0 & 5.2 & 0.58 \\
\hline \rule{0pt}{2.5ex}\text{Qualitative Comp.} \rule[-1ex]{0pt}{0pt}& \text{Larger} & \text{Larger} & \text{Slower} \\
\hline \end{array}

 
Comparing radii:

\(\dfrac{r_w}{r_v}=\dfrac{10}{3.3}=3\)
 

As moons \(V\) and \(W\) are both orbiting the same body, the ratio \(\dfrac{r^3}{T^2}\) will be the same for both.

Comparing orbital periods:

\(\dfrac{r_w^3}{T_w^2}\) \(=\frac{r_v^3}{T_v^2}\)  
\(\left(\dfrac{T_w}{T_v}\right)^2\) \(=\left(\dfrac{r_w}{r_v}\right)^3=3^3\)  
\(\dfrac{T_w}{T_v}=\) \(=\sqrt{27}=5.2\)  

 
Comparing orbital velocities.

\(v=\dfrac{2 \pi r}{T}\)

 
Using the calculations for radius and period:

\(v_w\) \(=\dfrac{2 \pi r_w}{T_w}\)  
  \(=\dfrac{2 \pi\left(3.0 \times r_v\right)}{5.2 \times T_v}\)  
  \(=\dfrac{3.0}{5.2} \times \dfrac{2 \pi r_v}{T_v}\)  
  \(=\dfrac{3.0}{5.2} \times v_v\)  
\(\dfrac{v_w}{v_v}\) \(=0.58\)  

Show Worked Solution

\begin{array} {|l|c|c|c|}
\hline  \rule{0pt}{2.5ex}  & \textit{Orbital radius} & \textit{Orbital period} & \textit{Orbital velocity} \\
\hline \rule{0pt}{2.5ex}\text{Quantitative Comp.} \rule[-1ex]{0pt}{0pt}& 3.0 & 5.2 & 0.58 \\
\hline \rule{0pt}{2.5ex}\text{Qualitative Comp.} \rule[-1ex]{0pt}{0pt}& \text{Larger} & \text{Larger} & \text{Slower} \\
\hline \end{array}

 
Comparing radii:

\(\dfrac{r_w}{r_v}=\dfrac{10}{3.3}=3\)
 

As moons \(V\) and \(W\) are both orbiting the same body, the ratio \(\dfrac{r^3}{T^2}\) will be the same for both.

Comparing orbital periods:

\(\dfrac{r_w^3}{T_w^2}\) \(=\frac{r_v^3}{T_v^2}\)  
\(\left(\dfrac{T_w}{T_v}\right)^2\) \(=\left(\dfrac{r_w}{r_v}\right)^3=3^3\)  
\(\dfrac{T_w}{T_v}=\) \(=\sqrt{27}=5.2\)  

 
Comparing orbital velocities.

\(v=\dfrac{2 \pi r}{T}\)

 
Using the calculations for radius and period:

\(v_w\) \(=\dfrac{2 \pi r_w}{T_w}\)  
  \(=\dfrac{2 \pi\left(3.0 \times r_v\right)}{5.2 \times T_v}\)  
  \(=\dfrac{3.0}{5.2} \times \dfrac{2 \pi r_v}{T_v}\)  
  \(=\dfrac{3.0}{5.2} \times v_v\)  
\(\dfrac{v_w}{v_v}\) \(=0.58\)  

PHYSICS, M5 EQ-Bank 17 MC

Two identical masses are placed at points `P` and `Q`. The escape velocity and circular orbital velocity of the mass at point `P` are `v_{P_(esc)}` and `v_{P_{o rb}}`. The escape velocity and circular orbital velocity of the mass at point `Q` are `v_{Q_(e s c)}` and `v_{Q_{o rb}}`. The diagram is drawn to scale and `X` denotes the centre of Earth.
 

The velocity for a body in circular orbit is given by  `v_{o rb} = sqrt((GM)/r`.

What is the value of `(v_{Q_(e s c)})/v_{P_{o rb}}`?

  1. `0.5`
  2. `1`
  3. `sqrt(2)`
  4. `2`
Show Answers Only

`B`

Show Worked Solution

→ The escape velocity of an object is given by `v_(esc)=sqrt((2GM)/(r))`

→ As the diagram is to scale, it can be measured that the distance from `Q` to `X` is twice that from `P` to `X`.

→ Let  `r` = distance from `P` to `X`  and  `2r` = distance from `Q` to `X`:

→ `v_(Q_(esc))=sqrt((2GM)/(2r))=sqrt((GM)/(r))`

→ `v_(p_(text{orb}))=sqrt((GM)/(r))` 

→ Hence, `v_(Q_(esc))=v_(P_(text{orb}))` → `(v_(Q_(esc)))/(v_(p_(text{orb})))=1`

`=>B`

PHYSICS, M5 EQ-Bank 15 MC

The table shows data about the solar system.

\begin{array} {|l|c|c|}
\hline  \rule{0pt}{2.5ex} \quad\textit{Planet} \quad  & \textit{Average distance} & \textit{Period (days)} \\
\quad\textit{} \quad  \rule[-1ex]{0pt}{0pt}& \textit{from the sun (AU)} & \textit{} \\
\hline \rule{0pt}{2.5ex} \text{Venus} \rule[-1ex]{0pt}{0pt}& 0.72 & 224.7 \\
\hline \rule{0pt}{2.5ex} \text{Earth} \rule[-1ex]{0pt}{0pt}& 1 & 365 \\
\hline \rule{0pt}{2.5ex} \text{Mars} \rule[-1ex]{0pt}{0pt}& 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. \(7.9 \times 10^2\) days
  2. \(1.5 \times 10^3\) days
  3. \(1.6 \times 10^3\) days
  4. \(6.6 \times 10^3\) days
Show Answers Only

\(B\)

Show Worked Solution

\(\rightarrow  \dfrac{r^3}{T^2}=\dfrac{G M}{4 \pi^2}\)

\(\rightarrow \text{Since all planets are orbiting the same central body with mass, \(M, G\) and \(4 \pi^2\) are constants.}\)

\(\rightarrow \text{The ratio \(\dfrac{r^3}{T^2}\) will be the same for all planets in the solar system.}\)

\(\begin{aligned}
\rule{0pt}{2.5ex} \dfrac{r^3}{T^2} \rule[-1ex]{0pt}{0pt}& =\dfrac{r_e^3}{T_e^2} \\
\rule{0pt}{2.5ex} \dfrac{4.5^3}{T^2} \rule[-1ex]{0pt}{0pt}& =\dfrac{1^3}{365^2} \\
\rule{0pt}{2.5ex} T^2 \rule[-1ex]{0pt}{0pt}& =\dfrac{365^2}{4.5^3} \\
\rule{0pt}{2.5ex} \therefore T \rule[-1ex]{0pt}{0pt}& =\sqrt{\frac{365^2}{4.5^3}} \\
\rule{0pt}{2.5ex} \rule[-1ex]{0pt}{0pt}& =1462 \ \text{days} \\
\rule{0pt}{2.5ex}  \rule[-1ex]{0pt}{0pt}& \approx 1.5 \times 10^3 \text {days}
\end{aligned}\)

\(\Rightarrow B\)

PHYSICS, M5 2015 HSC 26

Consider the following two models used to calculate the work done when a 300 kg satellite is taken from Earth's surface to an altitude of 200 km.

You may assume that the calculations are correct.
 

  1. What assumptions are made about Earth's gravitational field in models `X` and `Y` that lead to the different results shown?   (2 marks)
  1. Why do models `X` and `Y` produce results that, although different, are close in value?   (1 mark)
  1. Calculate the orbital velocity of the satellite in a circular orbit at the altitude of 200 km.   (3 marks)
Show Answers Only

a.  Model `X`:

→ Assumes Earth’s gravitational field strength remains constant moving upwards from the surface.

Model `Y`:

→ Assumes Earth’s gravitational field strength changes with altitude.

b.   Similarity of results due to:

→ The variation in gravitational field strength from Earth’s surface to an altitude of 200 km is minimal, so both models `X` and `Y` produce similar results.

c.   `v=7797  text{ms}^(-1)`

Show Worked Solution

a.   Model `X`:

→ Assumes Earth’s gravitational field strength remains constant moving upwards from the surface.

Model `Y`:

→ Assumes Earth’s gravitational field strength changes with altitude.


♦ Mean mark (a) 54%.

b.   Similarity of results due to:

→ The variation in gravitational field strength from Earth’s surface to an altitude of 200 km is minimal, so both models `X` and `Y` produce similar results.


♦♦ Mean mark (b) 38%.

c.    Centripetal force = force due to gravity:

`F_(c)` `=F_(g)`  
`(mv^2)/(r)` `=(GMm)/(r^2)`  
`:.v` `=sqrt((GM)/(r))`  
  `=sqrt((6.67 xx10^(-11)xx6xx10^(24))/(6.58 xx10^(6)))`  
  `=7797  text{ms}^(-1)`  

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 2022 HSC 13 MC

Two satellites share an orbit around a planet. One satellite has twice the mass of the other.
  


  

Which quantity would be different for the two satellites?

  1. Speed
  2. Momentum
  3. Orbital period
  4. Centripetal acceleration
Show Answers Only

`B`

Show Worked Solution

The orbital velocity, `v=sqrt((GM)/(r))` is independent of the mass of the orbiting body.

→ Both satellites have the same orbital velocity.

→ The momentum,  `p=mv` of the satellite with mass 2`m` is twice that of the satellite with mass `m`.

→ The two satellites have different momentum.

`=>B`

PHYSICS M5 2022 HSC 10 MC

The orbital velocity, `v`, of a satellite around a planet is given by  `v=sqrt((GM)/(r))`.

Which graph is consistent with this relationship?
 

  

Show Answers Only

`D`

Show Worked Solution

Squaring both sides of the given equation gives:

`v^2=(GM)/(r)`

→   `v^2 prop (1)/(r)`

→ A line of `v^2` plotted against `(1)/(r)` will yield a linear relationship.

`=>D`

PHYSICS, M5 2019 HSC 14 MC

A satellite in circular orbit at a distance `r` from the centre of Earth has an orbital velocity `v`.

If the distance was increased to `2r`, what would be the satellite's orbital velocity?

  1. `(v)/2`
  2. `0.7v`
  3. `1.4v`
  4. `2v`
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`B`

Show Worked Solution

The satellites centripetal force is provided by gravity:

`F_(c)` `=F_(g)`  
`(mv^2)/(r)` `=(GMm)/(r^2)`  
`v` `=sqrt((GM)/(r))`  
`∴ v` `prop (1)/(sqrt(r))`  

 
Hence, doubling the satellites radius changes its velocity by a factor of `(1)/(sqrt(2))`

`=>B`


♦ Mean mark 42%.

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

b.   `2.0 xx10^(7)\ \text{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)\ \text{m}`