Sheet 01 Newton’s Law Of Gravitation, Superposition Theorem.
An orange of mass \(0.25kg\) falls from a tree. What is the acceleration of the orange towards the earth? Also, calculate the acceleration of the earth towards orange. Given: \(M_{earth}=\left(5.983\times10^{24}\right)kg\), \(R_{earth}=\left(6.378\times10^{6}\right)m\) and \(G=\left(6.67\times10^{-11}\right)Nm^{2}kg^{-2}\).
Ans:\(a_{orange}=9.81\frac{m}{s^{2}}\) and \(a_{earth}=\left(4.099\times10^{-25}\right)\frac{m}{s^{2}}\).
A body weighing \(36kg\) on the surface of earth. How much will it weigh on the surface of Mars whose mass is \(\frac{1}{9}\) and radius \(\frac{1}{2}\) that of earth?
Ans: \(W’=16kg\).
The period of earth around the sun is 1 Years and mean orbital radius of earth’s orbit around sun is \(\left(1.5\times10^{8}\right)km\). Calculate: the mass of sun if \(G=\left(6.67\times10^{-11}\right)Nm^{2}kg^{-2}\).
Ans: \(M_{s}=\left(2.0\times10^{30}\right)kg\).
A mass \(M\) is broken into two parts of masses \(M_{1}\) and \(M_{2}\) related so that the force of gravitational attraction between the two parts is maximum?
Ans: \(M_{1}=M_{2}=\frac{M}{2}\).
Three equal masses of m kg each are fixed at the vertices of an equilateral triangle ABC. Find; (i) Force acting on a mass 2m placed at the centroid G of the triangle. (ii) What is the force if the mass at vertex A is doubled. Take:AG=BG=CG=1m. **[figure]**
Ans:(i) \(Zero\) and (ii) \(2Gm^{2}\).
A sphere of mass \(40kg\) is attracted by a second sphere of mass \(50kg\) with a force of \(0.02gwt\). How much is the distance between the two masses? Given: \(G=\left(6.67\times10^{-11}\right)Nm^{2}kg^{-2}\).
Ans: \(r=2.6cm\).
How far from the earth must a body be placed along a line joining the sun to the earth so that the resultant gravitational pull on the body due top earth and the sun is zero? Distance between the sun and the earth is \(\left(1.5\times10^{8}\right)km\). Given: \(\frac{M_{sun}}{M_{earth}}=\left(3.25\times10^{5}\right)\).
