Question 1
1
6 marks
Question 1(a)
1(a)
Define gravitational potential at a point.
Easystructured1 marks
Answer
work done in bringing unit mass from infinity (to the point) B1
Question 1(b)
1(b)
The gravitational potential \(\phi\) at distance r from point mass M is given by the expression where G is the gravitational constant. Explain the significance of the negative sign in this expression.
Mediumstructured2 marks
Answer
gravitational force is (always) attractive B1 either as r decreases, object/mass/body does work or work is done by masses as they come together B1
Question 1(d)
1(d)
The planet in (c) has mass M and diameter \(6.8 \times 10^{3} \mathrm{~km}\). The product G M for this planet is \(4.3 \times 10^{13} \mathrm{Nm}^{2} \mathrm{~kg}^{-1}\). A rock, initially at rest a long distance from the planet, accelerates towards the planet. Assuming that the planet has negligible atmosphere, calculate the speed of the rock as it hits the surface of the planet. speed = \(\mathrm{ms}^{-1}\)
Hardstructured3 marks
Answer
\(\frac{1}{2} m v^{2}=m \Delta \phi\) \(v^{2}=2 \times G M / r \quad\) C1 \(=\left(2 \times 4.3 \times 10^{13}\right) /\left(3.4 \times 10^{6}\right) \quad\) C1 \(v=5.0 \times 10^{3} \mathrm{~m} \mathrm{~s}^{-1} \quad\) A1 (Use of diameter instead of radius to give \(v=3.6 \times 10^{3} \mathrm{~ms}^{-1}\) scores 2 marks)
Question 1
1
Explain why the two separate electric fields have opposite signs. 2. On Fig. 4.2, sketch the variation with x of the combined electric field due to the \(\alpha\)-particle and the proton for values of x from \(4 \mu \mathrm{~m}\) to \(16 \mu \mathrm{~m}\).
structured3 marks
Question 1(a)
1(a)
Define gravitational potential at a point.
Mediumstructured2 marks
Answer
work done in moving unit mass from infinity (to the point) M1 A1
Question 1(b)
1(b)
The Moon may be considered to be an isolated sphere of radius \(1.74 \times 10^{3} \mathrm{~km}\) with its mass of \(7.35 \times 10^{22} \mathrm{~kg}\) concentrated at its centre.
structured3 marks
Question 1(b)(i)
1(b)(i)
A rock of mass 4.50 kg is situated on the surface of the Moon. Show that the change in gravitational potential energy of the rock in moving it from the Moon's surface to infinity is \(1.27 \times 10^{7} \mathrm{~J}\).
Mediumstructured1 marks
Answer
gravitational potential energy =G M m / x energy \(=\left(6.67 \times 10^{-11} \times 7.35 \times 10^{22} \times 4.5\right) /\left(1.74 \times 10^{6}\right)\) M1 energy \(=1.27 \times 10^{7} \mathrm{~J}\) A0
Question 1(b)(ii)
1(b)(ii)
The escape speed of the rock is the minimum speed that the rock must be given when it is on the Moon's surface so that it can escape to infinity. Use the answer in (i) to determine the escape speed. Explain your working. speed = \(\mathrm{ms}^{-1}\)
Mediumstructured2 marks
Answer
change in grav. potential energy = change in kinetic energy B1 \(\frac{1}{2} \times 4.5 \times v^{2}=1.27 \times 10^{7}\) \(v=2.4 \times 10^{3} \mathrm{~m} \mathrm{~s}^{-1}\)
Question 1(c)
1(c)
The Moon in (b) is assumed to be isolated in space. The Moon does, in fact, orbit the Earth. State and explain whether the minimum speed for the rock to reach the Earth from the surface of the Moon is different from the escape speed calculated in (b).
Hardstructured2 marks
Answer
Earth would attract the rock / potential at Earth('s surface) not zero / <0 / at Earth, potential due to Moon not zero escape speed would be lower
Question 1
1
The mass M of a spherical planet may be assumed to be a point mass at the centre of the planet.
structured3 marks
Question 1(b)
1(b)
A second stone, initially at rest at infinity, travels towards the planet, as illustrated in Fig.1.2. The stone does not hit the surface of the planet.
structured3 marks
Question 1(b)(i)
1(b)(i)
Determine, in terms of the gravitational constant G and the mass M of the planet, the speed \(V_{0}\) of the stone at a distance x from the centre of the planet. Explain your working. You may assume that the gravitational attraction on the stone is due only to the planet.
Hardstructured3 marks
Answer
kinetic energy increase/change = loss/change in (gravitational) potential energy \(1 / 2 m V_{0}{ }^{2}=G M m / x\) \(V_{0}{ }^{2}=2 G M / x\) \(V_{0}=\sqrt{ }(2 G M / x)\) (max. 2 for use of r not x )
Question 1
1
+q,
structured1 marks
Question 1(a)
1(a)
Define gravitational potential at a point.
Mediumstructured2 marks
Answer
work done bringing unit mass from infinity (to the point)
Question 1(b)
1(b)
A stone of mass m has gravitational potential energy \(E_{\mathrm{P}}\) at a point X in a gravitational field. The magnitude of the gravitational potential at X is \(\phi\). State the relation between \(m, E_{\mathrm{P}}\) and \(\phi\).
Easystructured1 marks
Answer
\(E_{\mathrm{P}}=-m \phi\)
Question 1(c)
1(c)
An isolated spherical planet of radius R may be assumed to have all its mass concentrated at its centre. The gravitational potential at the surface of the planet is \(-6.30 \times 10^{7} \mathrm{Jkg}^{-1}\). A stone of mass 1.30 kg is travelling towards the planet such that its distance from the centre of the planet changes from 6 R to 5 R. Calculate the change in gravitational potential energy of the stone. change in energy = J
Mediumstructured4 marks
Answer
\(\phi \propto 1 / x\) either at 6 R from centre, potential is \(\left(6.3 \times 10^{7}\right) / 6 \quad\left(=1.05 \times 10^{7} \mathrm{~J} \mathrm{~kg}^{-1}\right)\) or change in potential \(=(1 / 5-1 / 6) \times\left(6.3 \times 10^{7}\right)\) change in energy \(=(1 / 5-1 / 6) \times\left(6.3 \times 10^{7}\right) \times 1.3\)
Question 1
1
An isolated spherical planet has a diameter of \(6.8 \times 10^{6} \mathrm{~m}\). Its mass of \(6.4 \times 10^{23} \mathrm{~kg}\) may be assumed to be a point mass at the centre of the planet.
structured4 marks
Question 1(c)
1(c)
A rock, initially at rest at infinity, moves towards the planet. At point P , its height above the surface of the planet is 3.5 D, where D is the diameter of the planet, as shown in Fig.1.1. Calculate the speed of the rock at point P , assuming that the change in gravitational potential energy is all transferred to kinetic energy. speed = \(\mathrm{ms}^{-1}\)
Mediumstructured4 marks
Answer
gravitational potential energy =(-) G M m / x \(v^{2}=2 G M / x\) \(x=4 D=4 \times 6.8 \times 10^{6}\) \(v^{2}=\left(2 \times 6.67 \times 10^{-11} \times 6.4 \times 10^{23}\right) /\left(4 \times 6.8 \times 10^{6}\right)\) \(=3.14 \times 10^{6}\) \(v=1.8 \times 10^{3} \mathrm{~m} \mathrm{~s}^{-1}\) (use of 3.5 D giving \(1.9 \times 10^{3} \mathrm{~ms}^{-1}\), allow max. 3)
Question 1
1
An isolated spherical planet has a diameter of \(6.8 \times 10^{6} \mathrm{~m}\). Its mass of \(6.4 \times 10^{23} \mathrm{~kg}\) may be assumed to be a point mass at the centre of the planet.
structured4 marks
Question 1(c)
1(c)
A rock, initially at rest at infinity, moves towards the planet. At point P , its height above the surface of the planet is 3.5 D, where D is the diameter of the planet, as shown in Fig.1.1. Calculate the speed of the rock at point P , assuming that the change in gravitational potential energy is all transferred to kinetic energy. speed = \(\mathrm{ms}^{-1}\)
Hardstructured4 marks
Answer
gravitational potential energy =(-) G M m / x \(v^{2}=2 G M / x\) C1 \(x=4 D=4 \times 6.8 \times 10^{6}\) \(v^{2}=\left(2 \times 6.67 \times 10^{-11} \times 6.4 \times 10^{23}\right) /\left(4 \times 6.8 \times 10^{6}\right)\) \(v=1.8 \times 10^{3} \mathrm{~m} \mathrm{~s}^{-1}\) (use of 3.5 D giving \(1.9 \times 10^{3} \mathrm{~ms}^{-1}\), allow max. 3)