State the unit and dimension of the universal gravitational constant (G)... | Physics

GRAVITATION

Very Short Question Answers

Q1. State the unit and dimension of the universal gravitational constant (G).
Ans.

S.I Unit- $\,\,Nm^{2}/kg^{2}$ Dimensional Formula :$\,\,[M^{-1}\,L^{3}\,T^{-2}\,]$

Q2. State the vector form of Newton's law of gravitation.
Ans.

Gravitational force $F=-G \frac {m_{1}m_{2}}{r^{2}}\hat{r} \:\:(or) \:\:F=-G \frac {m_{1}m_{2}}{r^{3}}\vec{r}$

Q3. If the gravitational force of Earth on the Moon is F. what is the gravitational force of moon on earth? Do these forces form an action-reaction pair?
Ans.

The gravitational force of the moon on the earth = -F
These two forces are equal and opposite direction. Hence, they form action and reaction pair

Q4. What would be the change in acceleration due to gravity (g) at the surface, if the radius of Earth decreases by 2% keeping the mass of Earth constant?
Ans.

$We\, know \, that\, \frac{\bigtriangleup R}{R}=-2%$

$Change\, in\, acceleration \,due\,to\,gravity\,$

$\frac{\bigtriangleup g}{g}=-2\times \frac{\bigtriangleup R}{R}=-2\times-2=4$

Q5. As we go from one planet to another, how will a) the mass and b) the weight of a body change?
Ans.

The mass of the body is always constant.
Weight of the body also changes.

Q6. Keeping the length of a simple pendulum constant, will the time period be the same on all planets? Support your answer with reason.
Ans.

$Time\,Period \,\: T \propto \frac{1} {\sqrt g} $

Hence, the gravitational force varies from planet to planet, so the time period also changes from planet to planet.

Q7. Give the equation for the value of g at a depth 'd' from the surface of Earth. What is the value of 'g at the centre of Earth?
Ans.

GRAVITATION

Q8. What are the factors that make 'g the least at the equator and maximum at the poles?
Ans.

At Equator: i) equatorial radius of the earth is maximum. ii) latitude, φ=0°
At Poles: i) polar radius of the earth is minimum ii) latitude φ=90°
(Or) i) shape of the earth ii) rotation of the earth.

Q9. Hydrogen is in abundance around the sun but not around earth. Explain.
Ans.

The escape velocity on the sun is greater than r.m.s velocity of hydrogen.But the escape velocity on earth is less than the r.m.s velocity of hydrogen. So, hydrogen is in abundance around the sun and less around the earth.

Q10. What is the time period of revolution of a geostationary satellite? Does it rotate from West to East or from East to West?.
Ans.

Time period of revolution of a geostationary satellite is 24 hours. It rotates from west to east.

Q11. What are polar satellites?
Ans.

A satellite that revolves in a polar orbit is called a polar satellite.A polar orbit passes over north and south poles of the earth.Time period (T) = 100 min.

Short Question Answers

Q1. State Kepler's laws of planetary motion

Ans. Law of orbits: All the planets moves in elliptical orbits with the sun situated at one of the foci.

kepler1

Law of areas: The line that joins any planet to the sun sweeps equal area in equal intervals of time.

kepler2

Law of periods: The square of the time period of revolution of a planet is proportional to the cube of the semi-major axis of the ellipse traced out by the planet. $$T^{2}\propto a^{3}$$

kepler3

Q2. Derive the relation between acceleration due to gravity(g) at the surface of a planet and Gravitational constant(G)

Ans. Consider a body of mass $ m $ situated on the surface of the planet of mass $M$and radius $R$. According Newton's Universal law of gravitation the force of attraction between mass and planet is

$$F=G\frac{Mm}{R^{2}}$$

Where force $F=mg$

$$mg=G\frac{Mm}{R^{2}}$$ $$\\ \therefore g=\frac{GM}{R^{2}}$$

The above equation shows the relation between acceleration due to gravity$(g) $at the surface of a planet and Gravitational constant$(G)$.

Q3. How does the acceleration due to gravity(g) change for the same and depth(d)

Ans. Let $'g'$ be the acceleration due to gravity on the surface of the earth, $\rho$ is density of the earth
$$g=\frac{GM_{e}}{R^{2}}=\frac{4}{3}\pi\rho GR$$

$g_{h}$ be the acceleration due to gravity at a height $'h'$

$$g_h=\frac{GM_e}{(R+h)^2}$$

$$\begin{aligned}
&\frac{g_{_h}}{g}=\frac{GM_{e}}{(R+h)^{2}}\frac{R^{2}}{GM_{e}} \\
&\frac{g_{_h}}{g}=\frac{R^{2}}{(R+h)^{2}} \\
&g_{_h}=g\left(1-\frac{2h}{R}\right) \\
\end{aligned}$$

and $g_d$ be the acceleration due to gravity in a depth $'d'$
$$\begin{gathered}
g_{_d}=\frac{4}{3}\pi\rho G(R-d) \\
\frac{g_{_d}}{g}=\frac{R-d}{R}=1-\frac{d}{R} \\
g_{_d}=g\left(1-{\frac{d}{R}}\right)
\end{gathered}$$

Therefore, if $h=d$ ,than $g_{_h}<g_{_d}$

If $d=R$ than $g_{_d}=0$ it means at center of theearth acceleration due to gravity is zero.

Q4. What is orbital velocity? Obtain an expression for it

Ans. Orbital Velocity: The horizontal velocity of an object to revolve around the planet in a. circular orbit is called Orbital Velocity,

Expression:Consider a satellite of mass $m$ revolving around the Earth at a height $h$. Let M be the. mass and $R$ be the radius of the Earth.Let' $V_0 $ be the orbital velocity on the surface oftheEarth The gravitational force between Earth and satellite provides necessary centripetal force

orbital

Earth's Gravitational force on the body

$$F=\frac{GMm}{(R+h)^{2}}$$While satellite moves in orbits the centripetal force $$F=\frac{mV_o^2}{(R+h)}$$ From the above equations
$$\begin{gathered}
\frac{mV_{o}^{2}}{(R+h)}=\frac{GMm}{(R+h)^{2}} \\
V_{o}^{2}=\frac{GM}{(R+h)} \\

\end{gathered}$$ Where $R >> h, \: GM=gR^2$ $$V_{o} =\sqrt{gR} $$

Q5. What is escape velocity? Obtain an expression for it.

Ans. Definition : The minimum velocity required for an object to escape from the gravitational. influence of the planet is called Escape Velocity

Expression for escape velocity :Consider a body of mass m thrown with a velocity $V_{e}$ from a planet of mass $M$ and radius $R$.

escape

Kinetic energy of theobject is equal to potential energy and it is in opposite direction.
$$\begin{gathered}
\frac{1}{2}mV_{e}^{2}=-\left(-\frac{GMm}{R}\right) \\
V_{e}=\sqrt{\frac{2GM}{R}} \\
V_{e}=\sqrt{2gR}
\end{gathered}$$

Earth escape velocity $V_{e}= 11. 2$ $km/ s$

Q6. What is a geostationary satellite? State its uses.

Ans. Defination:A satellite whose time period of rotation is $24$ hours and rotates along the direction of rotation of earth in equatorial plane is called geostationary satellite.

Uses:

Study the upper layers of atmosphere.
Forecast the changes in the atmosphere
Know the shape and size of the earth.
Identify theminerals and natural resourcespresent inside and on the surface of the earth
Transmit the T.V.programmes to distant places

Q7. If two places are at the same height from the mean sea level: One is a mountain and other is in air. At which place will 'g' be greater? State the reason for your answer.

Ans. Acceleration due to gravity $(g) $is more on mountain due to the presence of mass of mountain.

Q8. The weight of an object is more at the poles than at the equator. At which of these can we get more sugar for the same weight? State the reason for your answer.

Ans. Acceleration due to gravity $(g)$at pole is greater than at equator. Hence, more sugar is obtained due to at equator for same weight $(W=mg).$

Q9. If a nut becomes loose and gets detached from a satellite revolving around the earth.will it fall down to earth or will it revolve around earth? Give reasons for your answer.

Ans. It will revolves in the same orbit due to inertia of motion.

Q10. An object projected with a velocity greater than or equal to 11.2 km/s will not return to earth. Explain the reason.

Ans. The escape velocity of the earth is$ 11.2 km/s$. Hence, An object projected with a velocity greater than or equal to $11.2 km/s$ will not return to earth.

Essay Question Answers

Q1. Define gravitational potential energy and derive an expression for it associated with

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Q2. Two particles of masses $m_{1}$ and $m_{2}$ Derive an expression for the variation of acceleration due to gravity (a) above and (b) below the surface of the Earth.

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Q3. State Newton's Universal Law of Gravitation. Explain how the value of the Gravitational constant (G) can be determined by Cavendish method.

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Problems Question Answers

Q1. Two spherical balls each of mass 1 kg are placed 1 cm apart. Find the gravitational force of attraction between them.

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Q2. The mass of a ball is four times the mass of another ball. When these balls are separated by a distance of 10 cm. the force of gravitation between them is $6.67 \times 10^{7}$ N. Find the masses of the two balls.

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Q3. Three spherical balls of masses 1kg. 2kg and 3kg are placed at the corners of an equilateral triangle of side 1 m. Find the magnitude of gravitational force exerted by the 2kg and 3kg masses on the 1kg mass.

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Q4. At a certain height above the earth's surface, the acceleration due to gravity is 4% of its value at the surface of earth. Determine the height.

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Q5. A satellite is orbiting the earth at a height of 1000km. Find its orbital speed.

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Q6. A satellite orbits the earth at a height equal to the radius of earth. Find It's (i) orbital speed and (ii) Period of revolution..

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Q7. The gravitational force of attraction between two objects decreases by 36% when the distance between them is increased by 4 m. Find the original distance between them.

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Q8. Four identical masses of m are kept at the corners of a square of side a. Find the gravitational force exerted on one of the masses by the other masses.

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Q9. Two spherical balls of 1kg and 4kg are separated by a distance of 12 cm. Find the distance of a point from the 1 kg mass at which the gravitational force on any mass becomes zero.

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Q10. Three uniform spheres each of mass m and radius R are kept in such a way that each touches the other two. Find the magnitude of the gravitational force on any one of the spheres due to the other two.

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Q11. Two satellites are revolving round the earth at different heights. The ratio of their orbital speeds is 2:1. If one of them is at a height of 100 km. what is the height of the other satellite?

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Q12. A satellite is revolving round in a circular orbit with a speed of 8 km s' at a height where the value of acceleration due to gravity is 8 m s. How high is the satellite from the Earth's surface? ( Radius of planet = 6000 km)

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Q13. (a) Calculate the escape velocity of a body from the Earth's surface. (b) If the Earth were made of wood, its mass would be 10% of its current mass. What would be the escape velocity. If the Earth were made of wood?

Ans. GRAVITATION