Two bodies of masses 10 kg and 100 kg are separated by adistance of 2m ( G = 6.67 × 10-11 Nm2 kg-2). The gravitational potential at the mid point on the line joining the two is
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Solution
\(V_{g}=\frac{– 6.67\times 10^{-11}\times 10}{1}-\frac{6.67\times 10^{-11}\times 100}{1}\)
= –6.67 × 10-10– 6.67 × 10-9
= – 6.67 × 10-10 × 11 =–7.3 × 10-9 J/kg
Time period of a simple pendulum inside a satellite orbiting earth is
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Solution
Since T=2π\(\sqrt{\frac{l}{g}}\)
but inside the satellite g = 0
So T = ∞
The ratio of the radii of the planets R1 and R2 is k. The ratio of the acceleration due to gravity is r. The ratio of the escape velocities from them will be
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Solution
We know that,Ve=\(\sqrt{(2gR)}\)
\(∴\frac{(V_{e})p_{1}}{(V_{e})p_{2}}=\frac{\sqrt{(2g_{1}R_{1})}}{(2g_{2}R_{2})}=\sqrt{\left ( \frac{g_{1}}{g_{2}} \right )}\times \sqrt{\left ( \frac{R_{1}}{R_{2}} \right )}=\sqrt{Kr}\)
If ve and ve represent the escape velocity and or bital velocity of a satellite corresponding to acircular orbit of radius R, then
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Solution
Ve=\(\sqrt{2}V_{O}\) where ve and vO are the escape velocity and orbital velocity respectively.
The kinetic energy needed to project a body of mass m from the earth surface (radius R) to infinity is
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Solution
\(\frac{1}{2}mv_{ese}^{2}=\frac{1}{2}m(\sqrt{2gR})^{2}=mgr\)
The escape velocity of a body depends upon mass as
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Solution
\(V_{esc}=\sqrt{2gR}\),where R is radius of the planet.Hence escape velocity is independent of m.
The time period of a satellite of earth is 5 hours. If the separation between the earth and the satellite is increased to 4 times the previous value, the new time period will become
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Solution
\(T^{2}\alpha R^{3}\)
\(\Rightarrow \left ( \frac{T_{1}}{T_{2}} \right )^{2}=\left ( \frac{R_{1}}{R_{2}} \right )^{3}\)
\(\Rightarrow \left ( \frac{T_{1}}{T_{2}} \right )=\left ( \frac{R_{1}}{R_{2}} \right )^{\frac{3}{2}}=\left ( \frac{1}{4} \right )^{\frac{3}{2}}\)
\(\Rightarrow \frac{T_{2}}{T_{1}}=(4)^{\frac{3}{2}}=8\)
⇒T2=8×T1=8×5=40 hours
At sea level, a body will have minimum weight at
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Solution
At poles, the effect of rotation is zero and also the distance from the centre of earth is least.
Two satellites revolve round the earth with orbital radii 4R and 16R, if the time period of first satellite is T then that of the other is
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Solution
\(\frac{T_{1}}{T_{2}}=\left ( \frac{R_{1}}{R_{2}} \right )^{\frac{3}{2}}\Rightarrow \frac{T}{T_{2}}=\left ( \frac{4R}{16R} \right )^{\frac{3}{2}}\)
⇒T2=8T
A planet revolves in an elliptical or bit around the sun. The semi-major and semi-minor axes are a and b. Then the square of time period, T is directly proportional to
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Solution
T2α(major axis)3 ⇒ T3 α a3