IB Physics HL D.1 Gravitational fields questions

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Question 1

Paper 2Easy6 marks

A probe is outside a small spherical moon. Figure 1 shows the gravitational potential VgV_g as a function of distance rr from the moon’s centre. The positive radial direction is defined to be away from the moon. The radial component of gravitational field strength is related to the potential gradient by

gr=−ΔVgΔr.g_r=-\frac{\Delta V_g}{\Delta r}.
1.51.61.71.81.922.12.22.32.42.5-2.8-2.6-2.4-2.2-2-1.8-1.6-1.4r / (106 m)Vg / (106 J kg-1)
(a)

Read the graph and state VgV_g at r=2.0×106 mr=2.0\times10^6\ \mathrm{m} and r=2.5×106 mr=2.5\times10^6\ \mathrm{m}

[2]
(b)

Calculate the magnitude of the average radial gravitational field strength between these two distances.

[3]
(c)

State the direction of the gravitational field relative to the moon.

[1]

Question 2

Multiple choiceMedium1 mark

A satellite is at a distance r=2.0×107 mr=2.0\times10^7\text{ m} from the centre of a planet. The graph shows the planet’s gravitational field strength gg at selected distances rr What is the escape speed from this position?

0123405101520r/(107 m)g/(N kg-1)
Show answer and explanation

Answer: B. At r=2.0×107 mr=2.0\times10^7\text{ m} the graph gives g=5.0 N kg−1g=5.0\text{ N kg}^{-1} Since g=GM/r2g=GM/r^2 vesc=2GM/r=2gr=2(5.0)(2.0×107)=1.4×104 m s−1v_{\rm esc}=\sqrt{2GM/r}=\sqrt{2gr}=\sqrt{2(5.0)(2.0\times10^7)}=1.4\times10^4\text{ m s}^{-1}

A: uses the circular orbital speed gr\sqrt{gr} omitting the factor 2 under the square root.
C: assumes escape speed is twice the circular orbital speed, rather than 2\sqrt{2} times it.
D: reads g=20 N kg−1g=20\text{ N kg}^{-1} from the graph at r=1.0×107 mr=1.0\times10^7\text{ m} not the value at the satellite’s radius.

Remember: escape speed is 2\sqrt{2} times circular orbital speed at the same radius.

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