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Gravitation

Q: If R\$=\$radius of the earth and g\$=\$acceleration due to gravity on the surface of the earth, the acceleration due to gravity at a distance (r>R) from the centre of the earth is proportional to

For \$r>R\$, earth can be considered as a point mass.Therefore, \$ F = \dfrac{GMm}{r^2}\$From the equation, \$F\$ is inversely proportional to \$r^2\$.

Q: The height at which the value of acceleration due to gravity becomes \$50\$% of that at the surface of the earth (Radius of the earth \$ = 6400\$ km) is nearly

Acceleration due to gravity a height \$h\$ from earth's surface is given by:\$g_h = g(\dfrac {R}{R+h})^2\$Here, \$g_h = \dfrac{g}{2}\$\$\Rightarrow \dfrac {g}{2}=g(\dfrac {R}{R+h})^2\$\$\Rightarrow \dfrac {R+h}{R}=1.414\$\$\Rightarrow \dfrac {h}{R}=0.414\$\$\Rightarrow h=0.414\times 6400=2649.6\ km\$

Q: A missile is launched with a velocity less than the escape velocity. The sum of its kinetic and potential energies is

When a body is projected with an initial speed \$v_{i}\$ from a point at a distance \$R_{E}+h\$ from Earth's centre, the total energy is given by\$E = \dfrac {1}{2}m{v_{i}^{2}} - \dfrac {GmM_{E}}{(R_{E}+h)}\$The body escapes Earth's gravitation when the first term in the right hand side is equal or greater than the second term, that is the escape speed. When that speed is less than the escape speed, first term is smaller than the second term implies Gravitational potential energy is higher than the kinetic energy. Hence the total energy will be negative.

Q: If R\$=\$radius of the earth and g \$=\$acceleration due to gravity on the surface of the earth, the acceleration due to gravity at a distance (r<R) from the centre of the earth is proportional to

\$F(d) = GM'm/r^2\$\$ M' = Mr^3/R^3\$\$F(d)= GMmr/R^3\$\$r = R-d\$\$g(d) = GM(R-d)/R^3\$\$g(d) = g[(R-d)/R]\$\$g(d) = g(1-\dfrac {d}{R})\$Therefore, the gravitational field varies proportional to \$r\$.

Q: A man covers \$60m\$ distance in one minute on the surface of earth. The distance he will cover on the surface of moon in one minute is \$\left ( g_{m}= \dfrac{g_{e}}{6} \right )\$

As the man moves along the surface, there is no effect of gravity on his ability to move.Hence, he can move the same distance as of Earth on Moon too.

Q: The depth at which the value of \$g\$ becomes \$25\%\$ of that at the surface of the earth. \$(\$ Radius of the earth \$=\$ 6400km.\$)\$

The value of \$g\$ at a depth is given by:\$g_d = g(1-\dfrac {d}{R})\$Here, \$g_d = \dfrac{g}{4}\$\$\Rightarrow \dfrac {g}{4}=g(1- \dfrac {d}{R})\$\$\Rightarrow \dfrac {d}{R}=\dfrac {3}{4}\$\$\Rightarrow d = 0.75\times 6400 = 4800\ km\$

Q: A body has weight (W) on the ground. The work which must be done to lift it to a height equal to the radius of the earth, R, is

If acceleration due to gravity was constant, constant force W was to be applied. However, due to the decrease in g we need to apply less force as we go higher. Hence the work done is lesser.

Q: The ratio of the radius of a planet A to that of planet B is \$r\$. The ratio of accelerations due to gravity for the two planets is \$x\$. The ratio of the escape velocities from the two planets is :

Escape Velocity is given by: \$V_{e}=\sqrt{2gR}\$Given: \$ \dfrac{R_A}{R_B} = r\$ and \$\dfrac{g_A}{g_B} = x \$Taking ratio of the escape velocities of the \$2\$ planets:\$\dfrac{V_A}{V_B} = \sqrt{rx}\$

Q: The ratio of acceleration due to gravity at a depth \$h\$ below the surface of earth and at a height \$h\$ above the surface for \$h<<R\$

Acceleration due to gravity at a depth \$h\$ below the surface of earth is\$g_d = g(1 - \dfrac{h}{R})\$Acceleration due to gravity at a height \$h\$ above the surface of earth is\$g_h = g(1- \dfrac{2h}{R})\$where, \$g\$ is the acceleration due to gravity on the surface and \$R\$ is the radius of earth.So, \$\dfrac{g_d}{g_h} = \dfrac{(R - h)}{ (R - 2h)}\$Since \$R>> h\$We can conclude that this ratio is constant.

Q: Consider Earth to be a homogeneous sphere. Scientist A goes deep down in a mine and scientist B goes high up in a balloon. The gravitational field measured by

The variation of 'g' with altitude is: \$g' = g(1- \dfrac{2h}{R})\$The variation of 'g' with depth is: \$g' = g(1- \dfrac{d}{R})\$So, from the above equations we can deduce that if a person goes deep into a mine or to a height above the Earth's surface, the gravitational field decreases at different rates.

Physics
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JEE Mains NEET AIIMS Easy Qs Med Qs Hard Qs

If R $=$ radius of the earth and g $=$ acceleration due to gravity on the surface of the earth, the acceleration due to gravity at a distance (r>R) from the centre of the earth is proportional to

r

r^{2}

r^{- 2}

r^{- 1}

Medium

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The height at which the value of acceleration due to gravity becomes $50$ % of that at the surface of the earth (Radius of the earth $= 6400$ km) is nearly

2650

2430

2250

2350

Medium

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A missile is launched with a velocity less than the escape velocity. The sum of its kinetic and potential energies is

positive

negative

zero

may be positive or negative depending upon its initial velocity

Medium

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If R $=$ radius of the earth and g $=$ acceleration due to gravity on the surface of the earth, the acceleration due to gravity at a distance (r

r

r^{2}

r^{- 2}

r^{- 1}

Medium

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A man covers $60 m$ distance in one minute on the surface of earth. The distance he will cover on the surface of moon in one minute is $(g_{m} = \frac{g _{e}}{6})$

60 m

60 X 6 m

\frac{6 0}{6} m

60 m

Medium

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The depth at which the value of $g$ becomes $25 %$ of that at the surface of the earth. $($ Radius of the earth $=$ 6400km. $)$

4800 km

2400 km

3600 km

1200 km

Medium

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A body has weight (W) on the ground. The work which must be done to lift it to a height equal to the radius of the earth, R, is

equal to W

\times

greater than W

\times

less than W

\times

Cannot be determined

Medium

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The ratio of the radius of a planet A to that of planet B is $r$ . The ratio of accelerations due to gravity for the two planets is $x$ . The ratio of the escape velocities from the two planets is :

r x

\frac{r}{x}

r

\frac{x}{r}

Medium

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The ratio of acceleration due to gravity at a depth $h$ below the surface of earth and at a height $h$ above the surface for $h < < R$

is constant.

increases linearly with h.

varies parabolically with h.

decreases.

Medium

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Consider Earth to be a homogeneous sphere. Scientist A goes deep down in a mine and scientist B goes high up in a balloon. The gravitational field measured by

A goes on decreasing and that of B goes on increasing

B goes on decreasing and that of A goes on increasing

each decreases at the same rate

each decreases at different rates

Medium

View solution>

Gravitation

Physics 6470 Views

If R=radius of the earth and g=acceleration due to gravity on the surface of the earth, the acceleration due to gravity at a distance (r>R) from the centre of the earth is proportional to

The height at which the value of acceleration due to gravity becomes 50% of that at the surface of the earth (Radius of the earth =6400 km) is nearly

A missile is launched with a velocity less than the escape velocity. The sum of its kinetic and potential energies is

If R=radius of the earth and g =acceleration due to gravity on the surface of the earth, the acceleration due to gravity at a distance (r

A man covers 60m distance in one minute on the surface of earth. The distance he will cover on the surface of moon in one minute is (gm​=6ge​​)

The depth at which the value of g becomes 25% of that at the surface of the earth. ( Radius of the earth = 6400km.)

A body has weight (W) on the ground. The work which must be done to lift it to a height equal to the radius of the earth, R, is

The ratio of the radius of a planet A to that of planet B is r. The ratio of accelerations due to gravity for the two planets is x. The ratio of the escape velocities from the two planets is :

The ratio of acceleration due to gravity at a depth h below the surface of earth and at a height h above the surface for h<<R

Consider Earth to be a homogeneous sphere. Scientist A goes deep down in a mine and scientist B goes high up in a balloon. The gravitational field measured by

Physics
6470 Views

If R $=$ radius of the earth and g $=$ acceleration due to gravity on the surface of the earth, the acceleration due to gravity at a distance (r>R) from the centre of the earth is proportional to

The height at which the value of acceleration due to gravity becomes $50$ % of that at the surface of the earth (Radius of the earth $= 6400$ km) is nearly

If R $=$ radius of the earth and g $=$ acceleration due to gravity on the surface of the earth, the acceleration due to gravity at a distance (r

A man covers $60 m$ distance in one minute on the surface of earth. The distance he will cover on the surface of moon in one minute is $(g_{m} = \frac{g _{e}}{6})$

The depth at which the value of $g$ becomes $25 %$ of that at the surface of the earth. $($ Radius of the earth $=$ 6400km. $)$

The ratio of the radius of a planet A to that of planet B is $r$ . The ratio of accelerations due to gravity for the two planets is $x$ . The ratio of the escape velocities from the two planets is :

The ratio of acceleration due to gravity at a depth $h$ below the surface of earth and at a height $h$ above the surface for $h < < R$