CAIE A-Level Physics 13.2.2 Newton's Law of GravitationPractise calculating and rearranging F = Gm1m2/r² for point masses, interpreting inverse-square dependence, direction, units and unknown quantities.Syllabus2028–2030CoursePhysics 9702LevelA2
Exam pointscalculate gravitational force using F = Gm₁m₂/r² for two point massesrearrange the law to determine mass, separation or gravitational constant and use consistent unitsinterpret the direction and inverse-square dependence of the gravitational force
13.2.2—Newton’s law of gravitation F = Gm1m2 / r2 for the force between two point question 1[Maximum number: 1]An isolated uniform conducting sphere has mass M and charge Q.The gravitational field strength at the surface of the sphere is g.The electric field strength at the surface of the sphere is E.Show that the numerical value of α\alphaα is 1.35×1020 kg2C−21.35 \times 10^{20} \mathrm{~kg}^{2} \mathrm{C}^{-2}1.35×1020 kg2C−2.Show AnswerMark as masteredα=1/(4π×6.67×10−11×8.85×10−12)=1.35×1020( kg2C−2) or α=(8.99×109)/(6.67×10−11)=1.35×1020( kg2C−2)\begin{aligned} \alpha=1 /\left(4 \pi \times 6.67 \times 10^{-11} \times 8.85 \times 10^{-12}\right)=1.35 \times 10^{20}\left(\mathrm{~kg}^{2} \mathrm{C}^{-2}\right) \text { or } \alpha=\left(8.99 \times 10^{9}\right) /\left(6.67 \times 10^{-11}\right)=1.35 \times 10^{20}\left(\mathrm{~kg}^{2} \mathrm{C}^{-2}\right) \end{aligned}α=1/(4π×6.67×10−11×8.85×10−12)=1.35×1020( kg2C−2) or α=(8.99×109)/(6.67×10−11)=1.35×1020( kg2C−2)A1Add to Test