IB Physics SL D.1 Gravitational fields questions

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Question 1

Paper 2Medium12 marks

A probe is outside a spherical asteroid of uniform density. The asteroid has radius 4.0×105 m4.0\times10^5\ \mathrm{m} The probe has mass 800 kg800\ \mathrm{kg} Figure 1 shows the meaning of rr and measured values of the gravitational force FF on the probe at different distances from the asteroid’s centre.

ORPrMeasured gravitational force on the probe4567891002468101214r/(105 m)F/N

The gravitational constant is G=6.67×10−11 N m2 kg−2G=6.67\times10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}

(a)

State how the asteroid’s mass is treated when calculating the gravitational force on the probe.

[1]
(b)

Use Figure 1 to state the measured gravitational force when r=6.0×105 mr=6.0\times10^5\ \mathrm{m}

[1]
(c)

Use this measurement to estimate the mass of the asteroid.

[3]
(d)

Using your estimate from (c), calculate the gravitational force on the 800 kg800\ \mathrm{kg} probe when r=8.0×105 mr=8.0\times10^5\ \mathrm{m} State whether this result agrees with Figure 1.

[3]
(e)

At r=8.0×105 mr=8.0\times10^5\ \mathrm{m} a second probe has a mass of 1.6×103 kg1.6\times10^3\ \mathrm{kg} Determine the gravitational force on this probe.

[2]
(f)

A student suggests that the readings at r=6.0×105 mr=6.0\times10^5\ \mathrm{m} and r=8.0×105 mr=8.0\times10^5\ \mathrm{m} are consistent with F∝1/rF\propto 1/r Use these readings to assess the suggestion.

[2]

Question 2

Multiple choiceHard1 mark

A satellite moves in an elliptical orbit around a planet. Its closest and farthest distances from the planet are 1.0×107 m1.0\times10^7\text{ m} and 7.0×107 m7.0\times10^7\text{ m} A satellite in a circular orbit around the same planet has radius 2.0×107 m2.0\times10^7\text{ m} and period 4.0 d4.0\text{ d} Which pair gives the elliptical orbit’s period, followed by the ratio of its speed at closest approach to its speed at farthest distance?

Show answer and explanation

Answer: C. For the same planet, Kepler’s third law gives T2/a3=constantT^2/a^3=\text{constant} and Kepler’s second law gives equal swept areas in equal times.

  1. The semi-major axis is a=(1.0+7.0)×107/2=4.0×107 ma=(1.0+7.0)\times10^7/2=4.0\times10^7\text{ m} Thus T=4.0(4.0/2.0)3/2=11.3 dT=4.0(4.0/2.0)^{3/2}=11.3\text{ d} which rounds to 11 d11\text{ d}
  2. At closest and farthest approach, velocity is perpendicular to the radius, so equal area rates give rclosevclose=rfarvfarr_{\rm close}v_{\rm close}=r_{\rm far}v_{\rm far} Hence vclose/vfar=7v_{\rm close}/v_{\rm far}=7 A: omits the cube in Kepler’s third-law relation, using T/T0=a/a0T/T_0=\sqrt{a/a_0} B: reverses the speed ratio. D: uses T/T0=(a/a0)3T/T_0=(a/a_0)^3 omitting the square root. Remember: orbital speed is greater nearer the central body.

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