Date | May 2021 | Marks available | 3 | Reference code | 21M.2.HL.TZ1.2 |
Level | Higher level | Paper | Paper 2 | Time zone | 1 |
Command term | Show that | Question number | 2 | Adapted from | N/A |
Question
A planet is in a circular orbit around a star. The speed of the planet is constant. The following data are given:
Mass of planet =8.0×1024 kg
Mass of star =3.2×1030 kg
Distance from the star to the planet R =4.4×1010 m.
A spacecraft is to be launched from the surface of the planet to escape from the star system. The radius of the planet is 9.1 × 103 km.
Explain why a centripetal force is needed for the planet to be in a circular orbit.
Calculate the value of the centripetal force.
Show that the gravitational potential due to the planet and the star at the surface of the planet is about −5 × 109 J kg−1.
Estimate the escape speed of the spacecraft from the planet–star system.
Markscheme
«circular motion» involves a changing velocity ✓
«Tangential velocity» is «always» perpendicular to centripetal force/acceleration ✓
there must be a force/acceleration towards centre/star ✓
without a centripetal force the planet will move in a straight line ✓
F=(6.67×10-11)(8×1024)(3.2×1030)(4.4×1010)2=8.8×1023 «N» ✓
Vplanet = «−»(6.67×10-11)(8×1024)9.1×106=«−» 5.9 × 107 «J kg−1» ✓
Vstar = «−»(6.67×10-11)(3.2×1030)4.4×1010=«−» 4.9 × 109 «J kg−1» ✓
Vplanet + Vstar = «−» 4.9 «09» × 109 «J kg−1» ✓
Must see substitutions and not just equations.
use of vesc = √2V ✓
v = 9.91 × 104 «m s−1» ✓