· H2 Physics — Ten-Year Series, worked ← All 8 questions

2023 A-Level · Paper 2 · Question 3

Two circular paths: one that ignores the mass, and one that depends on it

Gravitational field · magnetic field 6 marks Geostationary orbit Charged particle in a magnetic field

Two circular motions with one idea in common. A geostationary satellite is held in its circle by gravity, and a chlorine ion is held in its circle by a magnetic field. Each half asks for a number and then for an explanation: does every satellite, or every ion, have to match? Below is the full solution, and beside it both circles: change the mass and watch what stays fixed.

The model, live

Gravitational / magnetic forceIon pathGeostationary

Part (a)(i)

2 marks

A 6800 kg6800\,\text{kg} satellite is in geostationary orbit 36 000 km36\,000\,\text{km} above the Earth's surface. Find the force keeping it in orbit, given the Earth's radius 6400 km6400\,\text{km} and mass 6.0×1024 kg6.0\times10^{24}\,\text{kg}.

Approach

The force keeping the satellite in orbit is the Earth's gravitational pull. Newton's law of gravitation uses the distance between the centres, so add the Earth's radius to the height first.

r=36 000+6400=42 400 km=4.24×107 mr = 36\,000 + 6400 = 42\,400\,\text{km} = 4.24\times10^{7}\,\text{m}
F=GMmr2=(6.67×10−11)(6.0×1024)(6800)(4.24×107)2=1.51×103 NF = \frac{GMm}{r^2} = \frac{(6.67\times10^{-11})(6.0\times10^{24})(6800)}{(4.24\times10^{7})^2} = 1.51\times10^{3}\,\text{N}

To check your answer:

Work it as a centripetal force instead. A geostationary satellite has period 24 h24\,\text{h}, so ω=2π86 400=7.27×10−5 rad s−1\omega = \dfrac{2\pi}{86\,400} = 7.27\times10^{-5}\,\text{rad s}^{-1} and mrω2mr\omega^2 =(6800)(4.24×107)(7.27×10−5)2= (6800)(4.24\times10^{7})(7.27\times10^{-5})^2 =1.52×103 N= 1.52\times10^{3}\,\text{N}. The two routes agree.

force =1.5×103 N= 1.5\times10^{3}\,\text{N}

Watch

Using 36 000 km36\,000\,\text{km} as rr gives 2.1×103 N2.1\times10^{3}\,\text{N}. The height is measured from the surface; the law needs the distance from the centre.

Part (a)(ii)

1 mark

Explain, using equations, whether every geostationary satellite must have a mass of 6800 kg6800\,\text{kg}.

Approach

Write the condition for the orbit and see which quantities survive. The satellite's mass appears on both sides and cancels.

Gravity provides the centripetal force:

GMmr2=mrω2⟹r3=GMω2\frac{GMm}{r^2} = mr\omega^2 \quad\Longrightarrow\quad r^3 = \frac{GM}{\omega^2}

A geostationary satellite must have ω\omega equal to the Earth's rotation, so rr is fixed. Nothing in this equation depends on mm.

No. From GMmr2=mrω2\dfrac{GMm}{r^2} = mr\omega^2, the satellite's mass mm cancels, giving r3=GMω2r^3 = \dfrac{GM}{\omega^2}. The radius depends only on MM and ω\omega, so a satellite of any mass can be in geostationary orbit.

Part (b)(i)

2 marks

A chlorine ion of charge −1e-1e and momentum 3.4×10−23 N s3.4\times10^{-23}\,\text{N s} enters a 1.8 mT1.8\,\text{mT} field at 90∘90^\circ to the field lines. Find the radius of its circular path.

Approach

The magnetic force BqvBqv is always perpendicular to the velocity, so it provides the centripetal force. The question gives momentum rather than mass and speed, so write the radius in terms of p=mvp = mv.

Bqv=mv2r⟹r=mvBq=pBqBqv = \frac{mv^2}{r} \quad\Longrightarrow\quad r = \frac{mv}{Bq} = \frac{p}{Bq}
r=3.4×10−23(1.8×10−3)(1.60×10−19)=0.118 mr = \frac{3.4\times10^{-23}}{(1.8\times10^{-3})(1.60\times10^{-19})} = 0.118\,\text{m}

To check your answer:

A chlorine-35 ion has mass about 35×1.66×10−27=5.8×10−26 kg35 \times 1.66\times10^{-27} = 5.8\times10^{-26}\,\text{kg}, so its speed is p/m≈590 m s−1p/m \approx 590\,\text{m s}^{-1}. Then r=mvBq=(5.8×10−26)(590)(1.8×10−3)(1.6×10−19)=0.12 mr = \dfrac{mv}{Bq} = \dfrac{(5.8\times10^{-26})(590)}{(1.8\times10^{-3})(1.6\times10^{-19})} = 0.12\,\text{m}.

radius =0.12 m= 0.12\,\text{m}

Watch

Use the magnitude of the charge, 1.60×10−19 C1.60\times10^{-19}\,\text{C}. The sign of the charge decides which way the ion curves, not how tightly.

Part (b)(ii)

1 mark

Explain whether every ion of charge −1e-1e must follow a path of the same radius in this field.

Approach

Look at r=pBqr = \dfrac{p}{Bq}. With BB and qq fixed, the radius still depends on the momentum, and here nothing cancels.

No. r=mvBqr = \dfrac{mv}{Bq}, so with the same BB and qq the radius is proportional to the momentum mvmv. Ions of charge −1e-1e with a different mass or a different speed follow circles of different radius.

Note

This is the contrast the question is built on. In the orbit the mass appears in both the force and the centripetal requirement, so it cancels. In the magnetic field the force BqvBqv does not depend on the mass, so the mass survives in r=mv/Bqr = mv/Bq.

What this question was really testing

  1. Centre-to-centre distance. Height above the surface plus the Earth's radius.
  2. A force that provides the centripetal force. Gravity in one case, BqvBqv in the other. Set it equal to mv2/rmv^2/r or mrω2mr\omega^2 and see what is left.
  3. Reading an equation for what cancels. The satellite's mass cancels; the ion's mass does not. "Use relevant equations" means show it.

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