Gravitation Test - 12

The acceleration due to gravity cannot be zero on the surface of earth because the centre of gravity is inside the earth.

Because the gravitational force on mercury is due to sun. So not similar condition for sun and Mercury.

Around centre of mass total torque is zero for earth-sun system.

If force is central then orbit is elliptical.

Chunks of matter is also like a planet and follows Kepler’s law.

Inertial mass can not depend on gravitational force.

(a) \(g'={g\over (1+{h\over2})^2}\) here g’ < g

h ⇒ altitude (Increasing)

(c) If h → latitude (Increasing)

then g’ > g

From question –

(a) \(F={GMm\over a^3}={mv^2\over a}\)

or \(v\propto {1\over a}\) \(\Big[v=\sqrt{GM\over a}\Big]\)

Time period, \(T={2\pi a\over v}\)

\(T={2\pi a^2\over \sqrt{GM}}\)

or \(T^2 \propto a^4\)

Hence, orbit will not be elliptical.

(c) Force, \(F=\Big({GM\over a^3}\Big)m=g'm\)

\(g'={GM\over a^3}\)    [g' = accelerationdue togravity]

As g' is constant,

\(\therefore\) path of projectile will be approximately parabolic

According to the problem G’ = 10 G,

then gravitational acceleration due to gravity

\(g'={G'M\over R^2}={10GM\over R^2}=10g\)

Now, Weight of person = mg' = m x 10g = 10 mg

Force on the man due to the sun

\(F={G'M_sm\over r^2}={G'M_sm\over 10r^2}\)     \(\Big(\because\, 1M_s={m_s\over10}\Big)\)

as r >> R so F will be very small so, the effect at the sun will be neglected. Due to this region gravity pull on the person will increase. Due to it, walking on ground would become more difficult. Critical velocity, \(v_c\) is proportional to \(g\).

as \(g'>g\)

\(\therefore v'_c>v_c\)

Hence, rain drops will fall much faster.

To overcome the increased gravitational force of the earth, the aeroplanes will have to travel much faster.

From electrostatics, force of attraction will be produced due to opposite charges, if the sun and the planets carries huge amount of opposite charges. Then, electrostatic force of attraction will be large, also gravitational force is also attractive in nature and both the forces will be added and both are radial in nature.

Both the forces obey inverse square law and are central forces. As both the forces are of same nature, kepler’s laws will be valid.

Gravitational force between earth and sun.

\(Fg=G\Big({m_s\times m_e\over r^2}\Big)\)

This force provides the necessary centripetal force for the circular orbit of the earth around the sun. As G decreases with time. The gravitational force \(F_g\) will weaken with time. As \(F_g\) is changing with time due to it, the earth will be going around the sun not strictly in closed orbit and radius also increases, since the attraction force is getting weaker Hence, after long time the earth will leave the solar-system.

Given:

\(F_1\) = -\(F_2\)

= -\({r_{12}\over r^3_{12}}GM_0^2\Big({m_1m_2\over M_0^2}\Big)^n\)

\(r_{12}=r_1-r_2\)

\(g={F\over mass}\)

= \({GM_0^2(m_1m_2)^n\over r^2_{12}(M_0)^{2n}}\times {1\over mass}\)

As g depends on position vector, it is different for different objects.

As g is variable, so proportionality constant will not be constant in Kepler’s III law.

Now, the nature of force is central force Hence, Kepler’s I and II law will be valid

Now,n = (-)ve

g = \({GM_0^2(m_1m_2)^n\over r^2_{12}(M_0)^{-2n}}\times {1\over mass}\)

g = \({GM_0^2\over r^2_{12}}\Big({M_0^2\over m_1m_2}\Big)^n \times {1\over mass}\)

\(M_0>m_1\) or \(m_2\)

\(\therefore\) g > 0, hence, object lighter than water will sink in water

For small objects, say of sizes less than 100 m placed in uniform gravitational field then centre of mass is very close with the centre of gravity at the body. But when the size of object increases, its weight changes and its C.M. and C.G. become far from each other. Like in the case of spherical ball, the CM and the CG are the same, but in case of Mount Everest, its CM is less a bit above its CG.

Total potential energy = \(-{GMm\over 2R}\)

Due to the atmospheric friction (viscous force) acting on satellite, energy decreases continuously, radius of the orbit or hight decreases gradually and the satellite spirals down with decreasing speed till it burns in the denser layers of the atmosphere.

A geostationary satellite revolves around the earth will the same angular velocity and in the same sense as done by the earth about its own axis. i.e., west-east direction. A polar satellite revolves around the earth's pole in north-south direction. It is independent of earth’s rotation.