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A-Level CAIE Physics 13 2 Gravitational Force Between Point Masses Question Bank

Practice A-Level CAIE Physics 13 2 Gravitational Force Between Point Masses questions by syllabus topic with past-paper context, marks, difficulty and question previews on Eduninja.

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Question 1

1

1 marks

Question 1(b)

1(b)

An isolated uniform conducting sphere has mass M and charge Q. The gravitational field strength at the surface of the sphere is g. The electric field strength at the surface of the sphere is E.

structured1 marks

Question 1(b)(ii)

1(b)(ii)

Show that the numerical value of \(\alpha\) is \(1.35 \times 10^{20} \mathrm{~kg}^{2} \mathrm{C}^{-2}\).

Mediumstructured1 marks

Answer

\[ \begin{aligned} \alpha=1 /\left(4 \pi \times 6.67 \times 10^{-11} \times 8.85 \times 10^{-12}\right)=1.35 \times 10^{20}\left(\mathrm{~kg}^{2} \mathrm{C}^{-2}\right) \text { or } \alpha=\left(8.99 \times 10^{9}\right) /\left(6.67 \times 10^{-11}\right)=1.35 \times 10^{20}\left(\mathrm{~kg}^{2} \mathrm{C}^{-2}\right) \end{aligned} \] A1

Question 1

1

2 marks

Question 1(b)

1(b)

The Earth may be considered to be an isolated sphere of radius R with its mass concentrated at its centre. The variation of the gravitational potential \(\phi\) with distance x from the centre of the Earth is shown in Fig. 1.1. The radius R of the Earth is \(6.4 \times 10^{6} \mathrm{~m}\).

structured2 marks

Question 1(b)(iii)

1(b)(iii)

In practice, the Earth is not an isolated sphere because it is orbited by the Moon, as illustrated in Fig. 1.2. initial path of meteorite The initial path of the meteorite is also shown. Suggest two changes to the motion of the meteorite caused by the Moon. 1. 2.

Mediumstructured2 marks

Answer

e.g. speed / velocity / acceleration would be greater deviates / bends from straight path (any sensible ideas, 1 each, max 2 )

Question 1

1

2.

structured2 marks

Question 1(b)

1(b)

The Earth may be considered to be an isolated sphere of radius R with its mass concentrated at its centre. The variation of the gravitational potential \(\phi\) with distance x from the centre of the Earth is shown in Fig. 1.1. The radius R of the Earth is \(6.4 \times 10^{6} \mathrm{~m}\).

structured2 marks

Question 1(b)(iii)

1(b)(iii)

In practice, the Earth is not an isolated sphere because it is orbited by the Moon, as illustrated in Fig. 1.2. initial path of meteorite The initial path of the meteorite is also shown. Suggest two changes to the motion of the meteorite caused by the Moon.

Mediumstructured2 marks

Answer

e.g. speed / velocity / acceleration would be greater deviates / bends from straight path (any sensible ideas, 1 each, max 2 )

Question 1

1

gel and soft tissue,

structured8 marks

Question 1(c)

1(c)

The Earth and the Moon may be considered to be spheres that are isolated in space with their masses concentrated at their centres. The masses of the Earth and the Moon are \(6.00 \times 10^{24} \mathrm{~kg}\) and \(7.40 \times 10^{22} \mathrm{~kg}\) respectively. The radius of the Earth is \(R_{\mathrm{E}}\) and the separation of the centres of the Earth and the Moon is \(60 R_{\mathrm{E}}\), as illustrated in Fig. 1.2.

structured8 marks

Question 1(c)(i)

1(c)(i)

Explain why there is a point between the Earth and the Moon at which the gravitational field strength is zero.

Mediumstructured2 marks

Answer

fields of Earth and Moon are in opposite directions M1 either resultant field found by subtraction of the field strength or any other sensible comment A1 so there is a point where it is zero A0 [2] (allow \(F_{\mathrm{E}}=-F_{\mathrm{M}}\) for 2 marks)

Question 1(c)(ii)

1(c)(ii)

Determine the distance, in terms of \(R_{\mathrm{E}}\), from the centre of the Earth at which the gravitational field strength is zero.

Hardstructured3 marks

Answer

\(G M_{\mathrm{E}} / x^{2}=G M_{\mathrm{M}} /(D-x)^{2} \quad \mathrm{C} 1 \left(6.0 \times 10^{24}\right) /\left(7.4 \times 10^{22}\right)=x^{2} /\left(60 R_{\mathrm{E}}-x\right)^{2} \quad \mathrm{C} 1 x=54 R_{\mathrm{E}}\) A1

Question 1(c)(iii)

1(c)(iii)

On the axes of Fig. 1.3, sketch a graph to show the variation of the gravitational field strength with position between the surface of the Earth and the surface of the Moon.

Hardstructured3 marks

Answer

graph: g=0 at least \(\frac{2}{3}\) distance to Moon B1 \(g_{\mathrm{E}}\) and \(g_{\mathrm{M}}\) in opposite directions M1 correct curvature (by eye) and \(g_{\mathrm{E}}>g_{\mathrm{M}}\) at surface A1 [3]

Question 1

1

A planet of mass m is in a circular orbit of radius r about the Sun of mass M, as illustrated in Fig. 1.1. The magnitude of the angular velocity and the period of revolution of the planet about the Sun are \(\omega\) and T respectively.

structured2 marks

Question 1(c)

1(c)

Data for the planets Venus and Neptune are given in Fig. 1.2. Assume that the orbits of both planets are circular.

structured2 marks

Question 1(c)(i)

1(c)(i)

Use the expression in (b) to calculate the value of T for Neptune. T= years

Mediumstructured2 marks

Answer

either \(T^{2}=(45 / 1.08)^{3} \times 0.615^{2}\) or \(T^{2}=0.30 \times 45^{3} \quad \mathrm{C} 1\) T=165 years A1

Question 1

1

3 marks

Question 1(a)

1(a)

Newton's law of gravitation applies to point masses.

structured3 marks

Question 1(a)(i)

1(a)(i)

State Newton's law of gravitation.

Easystructured2 marks

Answer

force proportional to product of masses B1 force inversely proportional to square of separation B1

Question 1(a)(ii)

1(a)(ii)

Explain why, although the planets and the Sun are not point masses, the law also applies to planets orbiting the Sun.

Mediumstructured1 marks

Answer

separation much greater than radius / diameter of Sun / planet B1

Question 1

1

3 marks

Question 1(c)

1(c)

Two protons are isolated in space. Their centres are separated by a distance R. Each proton may be considered to be a point mass with point charge. Determine the magnitude of the ratio > force between protons due to electric field force between protons due to gravitational field

Hardstructured3 marks

Answer

either ratio is \(Q_{1} Q_{2} / 4 \pi \varepsilon_{0} m_{1} m_{2} G\) C1 or \(\quad \begin{aligned} F_{\mathrm{E}}=2.30 \times 10^{-28} \times R^{-2} \quad \text { (C1) }\end{aligned}\)

Question 1

1

the mass of Mars, mass = kg

structured3 marks

Question 1(a)

1(a)

A moon is in a circular orbit of radius r about a planet. The angular speed of the moon in its orbit is \(\omega\). The planet and its moon may be considered to be point masses that are isolated in space. Show that r and \(\omega\) are related by the expression Explain your working.

Hardstructured3 marks

Answer

gravitational force provides the centripetal force \(G M m / r^{2}=m r \omega^{2}\) (must be in terms of \(\omega\) ) \(r^{3} \omega^{2}=G M\) and G M is a constant

Question 1(b)

1(b)

Phobos and Deimos are moons that are in circular orbits about the planet Mars. Data for Phobos and Deimos are shown in Fig. 1.1.

structured1 marks

Question 1(b)(i)

1(b)(i)

Use data from Fig. 1.1 to determine

Mediumstructured0 marks

Answer

1. for Phobos, \(\omega=2 \pi /(7.65 \times 3600)\) \(=2.28 \times 10^{-4} \mathrm{rad} \mathrm{s}^{-1}\) 2. \(\left(9.39 \times 10^{6}\right)^{3} \times\left(2.28 \times 10^{-4}\right)^{2}=\left(1.99 \times 10^{7}\right)^{3} \times \omega^{2}\) \(\omega=7.30 \times 10^{-5} \mathrm{rad} \mathrm{s}^{-1}\) \(T=2 \pi / \omega=2 \pi /\left(7.30 \times 10^{-5}\right)\) \(=8.6 \times 10^{4} \mathrm{~s}\) =23.6 hours

Question 1(b)(ii)

1(b)(ii)

The period of rotation of Mars about its axis is 24.6 hours. Deimos is in an equatorial orbit, orbiting in the same direction as the spin of Mars about its axis. Use your answer in (i) to comment on the orbit of Deimos.

Mediumstructured1 marks

Answer

either almost 'geostationary' or satellite would take a long time to cross the sky

Question 1

1

4 marks

Question 1(a)

1(a)

State Newton's law of gravitation.

Easystructured2 marks

Answer

force proportional to product of masses and inversely proportional to square of separation (do not allow square of distance/radius) M1 either point masses or separation @ size of masses A1

Question 1(b)

1(b)

The Earth and the Moon may be considered to be isolated in space with their masses concentrated at their centres. The orbit of the Moon around the Earth is circular with a radius of \(3.84 \times 10^{5} \mathrm{~km}\). The period of the orbit is 27.3 days. Show that

structured2 marks

Question 1(b)(ii)

1(b)(ii)

the mass of the Earth is \(6.0 \times 10^{24} \mathrm{~kg}\).

Easystructured2 marks

Answer

\(G M=r^{3} \omega^{2}\) or \(G M=v^{2} r \quad\) C1 \(M=\left(3.84 \times 10^{5} \times 10^{3}\right)^{3} \times\left(2.66 \times 10^{-6}\right)^{2} /\left(6.67 \times 10^{-11}\right) \quad\) M1 (special case: uses \(g=G M / r^{2}\) with \(g=9.81, r=6.4 \times 10^{6}\) scores max 1 mark)

Question 1(c)

1(c)

The mass of the Moon is \(7.4 \times 10^{22} \mathrm{~kg}\).

structured0 marks

Question 1(c)(i)

1(c)(i)

Using data from (b), determine the gravitational force between the Earth and the Moon.

Mediumstructured0 marks

Answer

grav. force \(=\left(6.0 \times 10^{24}\right) \times\left(7.4 \times 10^{22}\right) \times\left(6.67 \times 10^{-11}\right) /\left(3.84 \times 10^{8}\right)^{2} \quad\) C1

Question 1

1

kinetic energy, change in kinetic energy = J 2. gravitational potential energy. change in potential energy = J

structured2 marks

Question 1(a)

1(a)

State Newton's law of gravitation.

Easystructured2 marks

Answer

force is proportional to the product of the masses and inversely proportional to the square of the separation either point masses or separation >> size of masses

Question 1(b)

1(b)

A satellite of mass m is in a circular orbit of radius r about a planet of mass M. For this planet, the product G M is \(4.00 \times 10^{14} \mathrm{Nm}^{2} \mathrm{~kg}^{-1}\), where G is the gravitational constant. The planet may be assumed to be isolated in space.

structured4 marks

Question 1(b)(i)

1(b)(i)

By considering the gravitational force on the satellite and the centripetal force, show that the kinetic energy \(E_{\mathrm{K}}\) of the satellite is given by the expression

Mediumstructured2 marks

Answer

gravitational force provides the centripetal force \(m v^{2} / r=G M m / r^{2}\) and \(E_{\mathrm{K}}=1 / 2 m v^{2}\) hence \(E_{\mathrm{K}}=G M m / 2 r\)

Question 1(b)(ii)

1(b)(ii)

The satellite has mass 620 kg and is initially in a circular orbit of radius \(7.34 \times 10^{6} \mathrm{~m}\), as illustrated in Fig. 1.1. Resistive forces cause the satellite to move into a new orbit of radius \(7.30 \times 10^{6} \mathrm{~m}\). Determine, for the satellite, the change in

Mediumstructured0 marks

Answer

1. \(\Delta E_{\mathrm{K}}=1 / 2 \times 4.00 \times 10^{14} \times 620 \times\left(\left\{7.30 \times 10^{6}\right\}^{-1}-\left\{7.34 \times 10^{6}\right\}^{-1}\right)\) \(=9.26 \times 10^{7} \mathrm{~J}\) (ignore any sign in answer) (allow \(1.0 \times 10^{8} \mathrm{~J}\) if evidence that \(E_{\mathrm{K}}\) evaluated separately for each r ) 2. \(\Delta E_{\mathrm{P}}=4.00 \times 10^{14} \times 620 \times\left(\left\{7.30 \times 10^{6}\right\}^{-1}-\left\{7.34 \times 10^{6}\right\}^{-1}\right)\) \(=1.85 \times 10^{8} \mathrm{~J}\) (ignore any sign in answer) (allow 1.8 or \(1.9 \times 10^{8} \mathrm{~J}\) )

Question 1(b)(iii)

1(b)(iii)

Use your answers in (ii) to explain whether the linear speed of the satellite increases, decreases or remains unchanged when the radius of the orbit decreases.

Mediumstructured2 marks

Answer

either \(\left(7.30 \times 10^{6}\right)^{-1}-\left(7.34 \times 10^{6}\right)^{-1}\) or \(\Delta E_{\mathrm{K}}\) is positive \(/ \mathrm{E}_{\mathrm{K}}\) increased speed has increased 2 (a)