IBDP Physics- D.1 Gravitational fields- IB Style Questions For HL Paper 2 -FA 2025
Question
Orbital period of Venus \(= 225\) days
Orbital period of Earth \(= 365\) days
Most-appropriate topic code (IB Physics):
▶️ Answer/Explanation
(a)
From Kepler’s third law, \(T^{2} \propto r^{3}\). Hence, \[ \frac{r_{V}^{3}}{r_{E}^{3}} = \frac{T_{V}^{2}}{T_{E}^{2}} \] so \[ \frac{r_{V}}{r_{E}} = \left(\frac{T_{V}}{T_{E}}\right)^{2/3} = \left(\frac{225}{365}\right)^{2/3} \approx 0.72 . \]
(b)
The gravitational attraction of the Sun supplies the centripetal force that keeps a planet in orbit. The orbital radius \(r\) and period \(T\) can be measured observationally. Equating gravitational and centripetal forces, \[ \frac{GMm}{r^{2}} = \frac{mv^{2}}{r} = m\left(\frac{4\pi^{2}}{T^{2}}\right)r . \] Rearranging gives \[ M = \frac{4\pi^{2}r^{3}}{GT^{2}} . \] Since \(G\) is known, measuring \(r\) and \(T\) allows the mass of the Sun to be determined.
(c)
The gravitational field strength is the rate of change of gravitational potential with distance: \[ g = \frac{\Delta V}{\Delta r} . \] Using the given values, \[ g = \frac{3.0 \times 10^{7}}{5.0 \times 10^{9}} = 6.0 \times 10^{-3}\,m\,s^{-2} \] (or \(6.0 \times 10^{-3}\,N\,kg^{-1}\)).
