Two bodies of masses \(m\) and \(M\) are placed at distance \(d\) apart. What is the gravitational potential \(V\) at the position where the gravitational field due to them is zero?
1. \(V = -\frac{G}{d}(m + M)\)
2. \(V = -\frac{G}{d} m\)
3. \(V = -\frac{GM}{d}\)
4. \(V = -\frac{G}{d}(\sqrt{m} + \sqrt{M})^2\)
View Answer
At the point where the field is zero, \(\frac{Gm}{r_1^2} = \frac{GM}{r_2^2}\), which gives \(r_1 = \frac{\sqrt{m}}{\sqrt{m} + \sqrt{M}} d\) and \(r_2 = \frac{\sqrt{M}}{\sqrt{m} + \sqrt{M}} d\). The potential is \(V = -\frac{Gm}{r_1} - \frac{GM}{r_2} = -\frac{G}{d}(\sqrt{m} + \sqrt{M})^2\).
Gravitational potential difference between a point on surface of planet and another point \(10\text{ m}\) above is \(4\text{ J/kg}\). Considering gravitational field to be uniform, how much work is done in moving a mass of \(2.0\text{ kg}\) from the surface to a point \(5.0\text{ m}\) above the surface?
1. 0.40 J
2. 2.5 J
3. 4.0 J
4. 8.0 J
View Answer
For a uniform field, potential difference is proportional to distance. Thus, \(\Delta V' = \frac{5}{10} \times 4 = 2\text{ J/kg}\). The work done is \(W = m \Delta V' = 2.0 \times 2 = 4.0\text{ J}\).
A satellite is seen after each 8 hours over equator at a place on the earth when its sense of rotation is opposite to the earth. The time interval after which it can be seen at the same place when the sense of rotation of earth & satellite is same will be :
1. 8 hours
2. 12 hours
3. 24 hours
4. 6 hours
View Answer
When rotating oppositely, \(\frac{1}{T_{\text{rel}}} = \frac{1}{T_s} + \frac{1}{T_e} \Rightarrow \frac{1}{8} = \frac{1}{T_s} + \frac{1}{24}\), which gives \(T_s = 12\text{ hours}\). When rotating in the same direction, \(\frac{1}{T_{\text{rel}}'} = \frac{1}{T_s} - \frac{1}{T_e} = \frac{1}{12} - \frac{1}{24} = \frac{1}{24}\), so \(T_{\text{rel}}' = 24\text{ hours}\).
If the earth be at one half its present distance from the sun, number of days in the year will be nearly
1. 129
2. 30
3. 200
4. 60
View Answer
According to Kepler's Third Law, \(T^2 \propto R^3\). Thus, \(\left(\frac{T'}{T}\right)^2 = \left(\frac{R'}{R}\right)^3 = \left(\frac{1}{2}\right)^3 = \frac{1}{8}\), which gives \(T' = \frac{365}{\sqrt{8}} \approx 129\text{ days}\).
A satellite revolves around a planet in an elliptical orbit of minor and major axes \(a\) and \(b\) respectively. If T be the time period of the satellite, then \(T^2\) is proportional to
1. \(\left(\frac{a+b}{2}\right)^3\)
2. \(\left(\frac{a-b}{2}\right)^3\)
3. \(a^3\)
4. \(b^3\)
View Answer
According to Kepler's Third Law, \(T^2\) is proportional to the cube of the semi-major axis. Since the major axis is given as \(b\), the semi-major axis is \(b/2\), making \(T^2 \propto b^3\).
Imagine a light planet revolving around a very massive star in a circular orbit of radius \(r\) with a period of revolution T. If the gravitational force of attraction between the planet and the star is proportional to \(r^{-5/2}\), then the square of the time period will be proportional to
1. \(r^3\)
2. \(r^2\)
3. \(r^{2.5}\)
4. \(r^{3.5}\)
View Answer
The centripetal force is \(F = m\omega^2 r = m\frac{4\pi^2}{T^2} r \propto \frac{r}{T^2}\). Given \(F \propto r^{-5/2}\), we get \(\frac{r}{T^2} \propto r^{-5/2} \Rightarrow T^2 \propto r^{3.5}\).