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PHYSICS, M5 2024 HSC 31

In a thought experiment, a projectile is launched vertically from Earth's surface. Its initial velocity is less than the escape velocity.

The behaviour of the projectile can be analysed by using two different models, Model \(A\) and Model \(B\) as shown.
  

The effects of Earth's atmosphere and Earth's rotational and orbital motions can be ignored.

Compare the maximum height reached by the projectile, using each model. In your answer, describe the energy changes of the projectile.   (4 marks)

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  • Both models start with identical kinetic energy, which completely transforms into gravitational potential energy when the object reaches its highest point.
  • In Model \(A\), a consistent amount of kinetic energy converts to gravitational potential energy for each meter of upward movement.
  • In Model \(B\), the gravitational field gets weaker at higher altitudes. As a result, less kinetic energy is converted to gravitational potential energy per meter as the object goes up.
  • The object in Model \(B\) will therefore climb higher than the object in Model \(A\) before all its kinetic energy is converted to potential energy.

Show Worked Solution

  • Both models start with identical kinetic energy, which completely transforms into gravitational potential energy when the object reaches its highest point.
  • In Model \(A\), a consistent amount of kinetic energy converts to gravitational potential energy for each meter of upward movement.
  • In Model \(B\), the gravitational field gets weaker at higher altitudes. As a result, less kinetic energy is converted to gravitational potential energy per meter as the object goes up.
  • The object in Model \(B\) will therefore climb higher than the object in Model \(A\) before all its kinetic energy is converted to potential energy.
♦ Mean mark 47%.

PHYSICS, M5 2019 VCE 4 MC

The magnitude of the acceleration due to gravity at Earth's surface is \(g\).

Planet \(\text{Y}\) has twice the mass and half the radius of Earth. Both planets are modelled as uniform spheres.

Which one of the following best gives the magnitude of the acceleration due to gravity on the surface of Planet \(\text{Y}\)?

  1. \(\dfrac{1}{2} g\)
  2. \(1g\)
  3. \(4g\)
  4. \(8g\)
Show Answers Only

\(D\)

Show Worked Solution

\(g=\dfrac{GM}{r^2}\)

\(\text{Planet Y:}\ \dfrac{G \times 2M}{(\frac{r}{2})^2}=\dfrac{2GM}{\frac{r^2}{4}}=8 \times \dfrac{GM}{r^2}=8g\)

\(\Rightarrow D\)

Mean mark 58%.

PHYSICS, M5 2020 VCE 2*

Jupiter's moon Ganymede is its largest satellite.

Ganymede has a mass of 1.5 × 10\(^{23}\) kg and a radius of 2.6 × 10\(^6\) m.

Determine the magnitude of Ganymede's surface gravity?   (2 marks)

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\(1.5\ \text{ms}^{-2}\)

Show Worked Solution
\(F\) \(=\dfrac{GMm}{r^2}\)  
\(mg\) \(=\dfrac{GMm}{r^2}\)  
\(g\) \(=\dfrac{GM}{r^2}\)  
  \(=\dfrac{6.67 \times 10^{-11} \times 1.5 \times 10^{23}}{(2.6 \times 10^6)^2}\)  
  \(=1.5\ \text{ms}^{-2}\)  

PHYSICS, M5 2021 VCE 4*

The planet Phobetor has a mass four times that of Earth. Acceleration due to gravity on the surface of Phobetor is 18 m s\(^{-2}\).

If Earth has a radius \(R\), calculate the radius of Phobetor in terms of \(R\)?   (2 marks)

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\(\sqrt{2.22} \times R\)

Show Worked Solution

\(g=\dfrac{GM}{R^2}\ \Rightarrow \  R=\sqrt{\dfrac{GM}{g}}\)

\(\text{Phobetor:}\ M → 4M\ \ \text{and}\ \ g → 1.8g\)

\(\text{Radius of Phobetor}\) \(=\sqrt{\dfrac{4GM}{1.8g}}\)  
  \(=\sqrt{\dfrac{2.22GM}{g}}\)  
  \(=\sqrt{2.22} \times \sqrt{\dfrac{GM}{g}}\)  
  \(=\sqrt{2.22} \times R\)  
♦ Mean mark 44%.

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)

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

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

\(\therefore\) Decreasing the orbital radius will also decrease the orbital period.

PHYSICS, M5 2023 VCE 3 MC

Space scientists want to place a satellite into a circular orbit where the gravitational field strength of Earth is half of its value at Earth's surface.

Which one of the following expressions best represents the altitude of this orbit above Earth's surface, where \(R\) is the radius of Earth?

  1. \(\dfrac{\sqrt{2} R}{2}-R\)
  2. \(\sqrt{2} R\)
  3. \((\sqrt{2} R)-R\)
  4. \(2 R-\sqrt{2} R\)
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\(C\)

Show Worked Solution
  • For gravitational field strength, \(r \propto \sqrt{\dfrac{1}{g}}\)
  • Therefore if \(g\) → \(\dfrac{g}{2}\), the radius will become \(\sqrt{2}R\).
  • Thus the altitude above the surface of the earth is \(\sqrt{2}R\ -\ R\).

\(\Rightarrow C\)

♦♦ Mean mark 39%.

PHYSICS, M5 2023 HSC 1 MC

The gravitational field strength acting on a spacecraft decreases as its altitude increases.

This is due to a change in the

  1. mass of Earth.
  2. mass of the spacecraft.
  3. density of the atmosphere.
  4. distance of the spacecraft from Earth's centre.
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\(D\)  

Show Worked Solution

\(g=\dfrac{GM}{r^2}\ \ \Rightarrow\ \ g \propto \dfrac{1}{r^2}\)

\(\therefore\) As r increases, the gravitational field strength decreases

\( \Rightarrow D\)

PHYSICS, M5 EQ-Bank 21

Planet `X` has a mass eight times that of Earth. The acceleration due to gravity on the surface of this planet is half that on the surface of Earth.

If Earth has a radius of 1, what is the radius of Planet `X` ?   (3 marks)

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`4`

Show Worked Solution

Radius of Earth:

`F_(g)` `=(GM_em)/(r_e)^2`  
`g` `=(F)/(m)=(GM_e)/(r_e)^2`  
`:.r_(e)` `=sqrt((GM_e)/(g))`  

 
Radius of planet `X`:

`(g)/(2)` `=(G xxM_x)/(r_x)^2`  
  `=(G xx8M_e)/(r_x)^2`  
`:.r_(x)` `=sqrt((16GM_e)/(g))=4r_e`  

 
`:.` Since Earth has a radius of 1 `=>\ r_(x)=4`

PHYSICS, M5 2018 HSC 7 MC

A planet `X` has twice the mass and twice the radius of Earth.

What is the magnitude of the gravitational acceleration close to the surface of planet `X`?

  1. `(1)/(2)  g`
  2. `1\ g`
  3. `2\ g`
  4. `4\ g`
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`A`

Show Worked Solution

Suppose Earth has a mass of `M` and a radius of `r`:

`F=mg_1=(GMm)/(r^2)\ \ => \ \ g_1=(GM)/(r^2)`
 

If `X` has a mass of `2M` and a radius of `2r`:

`g_2` `=(2GM)/((2r)^2)`  
  `=(2GM)/(4r^2)`  
  `=(GM)/(2r^2)`  
  `=(g_1)/(2)`  

 
`=>A`


Mean mark 53%.