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Physics Case study (4 Marks)

Gravitation · UP Board (UPMSP) STD 11 Science (UPMSP - English Medium). 8 questions.

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Case study (4 Marks)

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Question 14 Marks
Read the passage given below and answer the following questions from 1 to 5. Cavendish’s Experiment The figure shows the schematic drawing of Cavendish’s experiment to determine the value of the gravitational constant. The bar AB has two small lead spheres attached at its ends. The bar is suspended from a rigid support by a fine wire. Two large lead spheres are brought close to the small ones but on opposite sides as shown. The value of G from this experiment came to be $6.67\times10^{-11}\frac{\text{N-m}^2}{\text{Kg}^2}$
  1. The big spheres attract the nearby small ones by a force which is:
  1. equal and opposite
  2. equal but in same direction
  3. unequal and opposite
  4. None of the above
  1. The net force on the bar is:
  1. non-zero
  2. zero
  3. Data insufficient
  4. None of these
  1. The net torque on the bar is:
  1. zero
  2. non-zero
  3. F times the length of the bar, where F is the force of attraction between a big sphere and its neighbouring
  4. Both (b) and (c)
  1. The torque produces twist in the suspended wire. The twisting stops when:
  1. restoring torque of the wire equals the gravitational torque
  2. restoring torque of the wire exceeds the gravitational torque
  3. the gravitational torque exceeds the restoring torque of the wire
  4. None of the above
  1. After Cavendish’s experiment, there have been given suggestions that the value of the gravitational constant G becomes smaller when considered over very large time period (in billions of years) in the future. If that happens, for our earth:
  1. nothing will change
  2. we will become hotter after billions of years
  3. we will be going around but not strictly in closed orbits
  4. None of the above
Answer
  1. (a) equal and opposite.
Explanation:
The force of attraction on small spheres due to big sphere are equal and opposite in direction. Hence, equal and opposite force separated by a fixed distance forms a couple.
  1. (b) zero
Explanation:
$\big|\text{F}_\text{net}\big|=\text{Zero}$
  1. (d) Both (b) and (c)
Explanation:
Magnitude of torque due to a couple
= (Either Force) × (Distance between of forces)
= F × l
where, l = length of the bar and F = force of attraction between a big sphere and its neighbouring small sphere.
  1. (a) restoring torque of the wire equals the gravitational torque.
Explanation:
The torque produces a twist in the suspended wire. The twisting stops when the restoring torque of the wire equal the gravitational torque.
  1. (c) we will be going around but not strictly in closed orbits
Explanation:
We know that, gravitational force between the earth and the sun.
$\text{F}_\text{g}=\frac{\text{GMm}}{\text{r}^2}$,
where M is mass of the sun and m is mass of the earth. When G decreases with time, the gravitational force $F_G$ will become weaker 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.
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Question 24 Marks
Read the passage given below and answer the following questions from 1 to 5. LAW OF ORBIT: The orbit of every planet is an ellipse around the sun with sun at one of the two foci of ellipse.
LAW OF AREAS: The line that joins a planet to the sun sweeps out equal areas in equal intervals of time. Area covered by the planet while revolving around the sun will be equal in equal intervals of time. This means the rate of change of area with time is constant. LAW OF PERIOD: According to this law the square of time period of a planet is directly proportional to the cube of the semi-major axis of its orbit. Suppose earth is revolving around the sun then the square of the time period (time taken to complete one revolution around sun) is directly proportional to the cube of the semi major axis. It is known as Law of Periods as it is dependent on the time period of planets. Answer the following.
  1. Keplers second law is knows as:
  1. Law of period
  2. Law of area
  3. Law of gravity
  4. None of these
  1. Keplers third law is knows as:
  1. Law of period
  2. Law of area
  3. Law of gravity
  4. None of these
  1. The velocity of a planet is constant throughout its elliptical trajectory in an orbit.
  1. True
  2. False
  3. None of these
  1. State Kepler’s second law of planetary motion.
  1. Two objects of masses $5kg$ and $10 kg$ separated by distance 10m. What is gravitational force between them?
Answer
  1. (c) Law of gravity
  1. (a) Law of period
  1. (a) True
  1. Keplers 3 laws are stated below.
LAW OF ORBIT: The orbit of every planet is an ellipse around the sun with sun at one of the two foci of ellipse.
LAW OF AREAS: The line that joins a planet to the sun sweeps out equal areas in equal intervals of time. Area covered by the planet while revolving around the sun will be equal in equal intervals of time. This means the rate of change of area with time is constant.
LAW OF PERIOD: According to this law the square of time period of a directly proportional to the cube of the semi-major axis of its orbit.
The force of attraction between any two unit masses separated by a unit distance is called universal gravitational constant denoted by G measured in $Nm^2/kg^2$.
  1. Mathematically,
$\text{F}=\text{G}\frac{\text{m}1\times\text{m}2}{\text{d}2}$
Here M1 = 5kg
M2 = 10kg
D = 10m
Then, forec is given by
$\text{F}=6.67\times10^{-11}\times5\times\frac{10}{100}$
$F = 3.33 \times 10^{-11}N$.
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Question 34 Marks
Read the passage given below and answer the following questions from 1 to 5. Acceleration due to gravity The acceleration for any object moving under the sole influence of gravity is known as acceleration due to gravity. So, for an object of mass m, the acceleration experienced by it is usually denoted by the symbol g which is related to F by Newton’s second law by relation F = mg Thus, $\text{g}=\frac{\text{F}}{\text{m}}=\frac{{\text{Gm}_\text{e}}}{\text{R}^2_\text{e}}$ Acceleration g is readily measurable as Re is a known quantity. The measurement ofG by Cavendish’s experiment (or otherwise), combined with knowledge of g and Re enables one to estimate $M_e$ from the above equation. This is the reason why there is a popular statement regarding Cavendish “Cavendish weighed the earth”. The value of g decrease as we go upwards from the earth’s surface or downwards, but it is maximum at its surface.
  1. If g is the acceleration due to gravity at the surface of the earth, the force acting on the particle of mass m placed at the surface is:
  1. mg
  2. $\frac{\text{GmM}\theta}{\text{R}^2_\text{e}}$
  3. Data insufficient
  4. Both (a) and (b)
  1. The weight of a body at the centre of earth is:
  1. same as on the surface of earth
  2. same as on the poles
  3. same as on the equator
  4. None of the above
  1. If the mass of the sun is ten times smaller and gravitational constant G is ten times larger in magnitude, then for earth:
  1. walking on ground would become more easy
  2. acceleration due to gravity on the earth will not change
  3. raindrops will fall much slower
  4. airplanes will have to travel much faster
  1. Suppose, the acceleration due to gravity at the earth’s surface is $10 ms^{-2}$ and at the surface of mars, it is $4.0 ms ^{-2}$ 60kg passenger goes from the earth to the mars in a spaceship moving with a constant velocity. Neglect all other objects in the sky. Which curve best represents the weight (net gravitational force) of the passenger as a function of time?
  1. A
  2. B
  3. C
  4. D
  1. If the mass of the earth is doubled and its radius halved, then new acceleration due to the gravity $\acute{\text{g}}$ is:
  1. $\acute{\text{g}}=4\text{g}$
  2. $\acute{\text{g}}=8\text{g}$
  3. $​​\acute{g}=\text{g}$
  4. $\acute{\text{g}}=16\text{g}$
Answer
  1. (d) Both (a) and (b)
Explanation:
The force acting on the particle of mass m at surface of the earth,
F = mg ...(i)
where, g = acceleration due to gravity at the earth’s surface.
Also, $\text{g}=\frac{\text{GM}_\text{e}}{\text{R}^2_\text{e}}$...(ii)
Then, from Eqs. (i) and (ii), we get
$\Rightarrow\text{F}=\text{mg}=\frac{\text{GmM}_\text{e}}{\text{R}^2_\text{e}}$
  1. (d) None of the above
Explanation:
Gravitational acceleration (g) at the centre of earth is zero, hence weight of body ( w = mg) at the centre of earth becomes zero.
  1. (d) airplanes will have to travel much faster
Explanation:
Consider the given diagram

Force on the object due to the earth,
$\text{F}=\frac{\acute{\text{G}\text{M}_\text{e}\text{m}}}{\text{R}^2}=\frac{10\text{GM}_\text{e}\text{M}}{\text{R}^2}(\because\acute{\text{g}}=10\text{G})$
$=10\bigg(\frac{\text{Gm}_\text{e}\text{m}}{\text{R}^2}\bigg)=(10\text{g})\text{m}=10\text{ mg}$ ...(i)
$\because\text{g}=\frac{\text{G}\text{M}_\text{e}}{\text{R}^2}$
Now, force on the object due to the sun,
$\acute{\text{F}}=\frac{{\text{G}\acute{\text{M}}_\text{s}\text{m}}}{\text{r}^2}$
$=\frac{\text{G}(\text{M}_\text{s})\text{m}}{10\text{ r}^2}$
$\big(\because\acute{\text{M}_\text{s}}=\frac{\text{M}_\text{s}}{10}\big)$
As, r R >> (radius of the earth)
⇒ F will be very small, so the effect of the sun will be neglected.
Now, as $\acute{\text{g}}=10\text{ g}$
Hence, weight of person $=\text{m}\acute{\text{g}}=10\text{ mg}$
$\brack{\text{from Eq.(i)}}$
i.e. Gravity pull on the person will increase. Due to it, walking on ground would become more difficult. Escape velocity $v_e$ is proportional to g, i.e.
$\text{V}_\text{e}\infty\text{ g}.$
As, $\acute{\text{g}}>\text{g}\Rightarrow\text{v}_\text{e}\acute{>}\text{v}_\text{e}$
Hence, rain drops will fall much faster. To overcome the increased gravitational force of the earth, the airplanes will have to travel much faster.
  1. (c)C
Explanation:
Initially, the weight of the passenger at the earth’s surface, w = mg = 60 × 10 = 600 N. Finally, the weight of the passenger at the surface of the mars = 60 × 4 = 240 N and during the flight in between somewhere its weight will be zero because at that point, gravitational pull of earth and mars will be equal.
  1. (b) $\acute{\text{g}}=8\text{ g}$
Explanation:
As we know that, acceleration due to gravity,
$\text{g}=\frac{\text{Gm}}{\text{R}^2}$
Given, $\acute{\text{M}}=2\text{M}(\because\text{Mass gets doubled)})$
$\Rightarrow\acute{\text{g}}=\frac{\text{G}\acute{\text{M}}}{\acute{\text{R}}^2}=\frac{\text{G}(2\text{M})}{(\frac{\text{R}}{2})^2}=\frac{8\text{ Gm}}{\text{R}^2}$
$\therefore\acute{\text{g}}=8\text{ g}$
Thus, the new acceleration due to gravity $\acute{\text{g}}$ is 8 times that of g.
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Question 44 Marks
Read the passage given below and answer the following questions from 1 to 5. We know that the earth attracts every object with a certain force and this force depends on the mass (m) of the object and the acceleration due to the gravity (g). The weight of an object is the force with which it is attracted towards the earth. Mathematically Where, W = weight of object m = mass of object g = acceleration due to the gravitational force As the weight of an object is the force with which it is attracted towards the earth, the SI unit of weight is the same as that of force, that is, Newton (N). The weight is a force acting vertically downwards; it has both magnitude and direction. We have learnt that the value of g is constant at a given place. Therefore at a given place, the weight of an object is directly proportional to the mass, say m, of the object, that is, W α m. It is due to this reason that at a given place, we can use the weight of an object as a measure of its mass. Answer the following questions.
  1. Dimensions of acceleration due to the gravity (g) is:
  1. $[ML^1 T^{-2}]$
  2. $[ML^{-1} T^{-2}]$
  3. $[ML^1 T^{-3}]$
  4. None of these
  1. SI unit of weight is same as:
  1. Force
  2. Mass
  3. Acceleration due to gravity
  4. None of these
  1. Which of the following has same unit?
  1. Mass and weight
  2. Weight and force
  3. Pressure and stress
  4. Both b and c
  1. Whether weight is scalar quantity or vector quantity? Justify your answer.
  1. Differentiate between mass and weight.
Answer
  1. (b) $[ML^{-1} T^{-2}]$
  1. (a) Force
  1. (b) Weight and force
  1. Weight is vector quantity as it has magnitude as well as direction which is always towards centre of a earth.
  1. Difference between mass and weight is given below
No
mass
weight
1
Mass is amount of matter in a body.
Weight is the measure of force acting on a mass due to acceleration due to gravity.
2
it is a scalar quantity
it is a vector quantity
3
SI unit of mass is Kilogram (Kg).
SI unit of weight is Newton (N).
4
Mass can never be zero
Weight can be zero where gravity is zero.
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Question 54 Marks
Read the passage given below and answer the following questions from 1 to 5. If a stone is thrown by hand, we see it falls back to the earth. Of course using machines we can shoot an object with much greater speeds and with greater and greater initial speed, the object scales higher and higher heights. A natural query that arises in our mind is the following: can we throw an object with such high initial speeds that it does not fall back to the earth ? Thus minimum speed required to throw object to infinity away from earth’s gravitational field is called escape velocity. $\text{V}_{\text{e}}=\sqrt{(2\text{gr})}$ Where g is acceleration due to gravity and r is radius of earth and after solving ve $11.2$ km/s. This is called the escape speed, sometimes loosely called the escape velocity. This applies equally well to an object thrown from the surface of the moon with g replaced by the acceleration due to Moon’s gravity on its surface and r replaced by the radius of the moon. Both are smaller than their values on earth and the escape speed for the moon turns out to be $2.3$ km/s, about five times smaller. This is the reason that moon has no atmosphere. Gas molecules if formed on the surface of the moon having velocities larger than this will escape the gravitational pull of the moon. Earth satellites are objects which revolve around the earth. Their motion is very similar to the motion of planets around the Sun and hence Kepler’s laws of planetary motion are equally applicable to them. In particular, their orbits around the earth are circular or elliptic. Moon is the only natural satellite of the earth with a near circular orbit with a time period of approximately $27.3$ days which is also roughly equal to the rotational period of the moon about its own axis.
  1. Time period of moon is:
  1. 27.3 days
  2. 20 days
  3. 85 days
  4. None of these
  1. Escape velocity from earth is given by:
  1. 20 km/s
  2. 11.2 km/s
  3. 2 km/s
  4. None of these
  1. Define escape velocity. Give its formula.
  1. Why moon don’t Have any atmosphere?
  1. What is satellite? Which law governs them?
Answer
  1. (a) 27.3 days
  1. (b) 11.2 km/s
  1. Minimum speed required to throw object to infinity away from earth’s gravitational field is called escape velocity.
$\text{V}_{\text{e}}=\sqrt{(2\text{gr})}$
Where g is acceleration due to gravity and r is radius of earth and after solving $v_e 11.2 km/s$. This is called the escape speed, sometimes loosely called the escape velocity.
  1. The escape speed for the moon turns out to be $2.3$ km/s, about five times smaller than that of earth. Therefore all atmospheric gas can go easily out of atmosphere of moon. This is the reason that moon has no atmosphere.
  1. Earth satellites are objects which revolve around the earth. Their motion is very similar to the motion of planets around the Sun and hence Kepler’s laws of planetary motion are equally applicable to them.
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Question 64 Marks
Read the passage given below and answer the following questions from 1 to 5. Satellites in a circular orbits around the earth in the equatorial plane with T = 24 hours are called Geostationary Satellites. Clearly, since the earth rotates with the same period, the satellite would appear fixed from any point on earth. It takes very powerful rockets to throw up a satellite to such large heights above the earth but this has been done in view of the several benefits of many practical applications. Thus radio waves broadcast from an antenna can be received at points far away where the direct wave fails to reach on account of the curvature of the earth. Waves used in television broadcast or other forms of communication have much higher frequencies and thus cannot be received beyond the line of sight. A Geostationery satellite, appearing fixed above the broadcasting station can however receive these signals and broadcast them back to a wide area on earth. The INSAT group of satellites sent up by India is one such group of geostationary satellites widely used for telecommunications in India. Another class of satellites is called the Polar satellites. These are low altitude (500 to 800 km) satellites, but they go around the poles of the earth in a north-south direction whereas the earth rotates around its axis in an east-west direction. Since its time period is around 100 minutes it crosses any altitude many times a day. However, since its height h above the earth is about 500-800 km, a camera fixed on it can view only small strips of the earth in one orbit. Adjacent strips are viewed in the next orbit, so that in effect the whole earth can be viewed strip by strip during the entire day. These satellites can view polar and equatorial regions. at close distances with good resolution. Information gathered from such satellites is extremely useful for remote sensing, meterology as well as for environmental studies of the earth.
  1. Time period of geospatial satellite is:
  1. 24 hours
  2. 48 hours
  3. 72 hours
  4. None of these
  1. Polar satellites are approximately revolving at height of
  1. 500 to 800km
  2. 1500 to 2000km
  3. 3000 to 4000km
  4. None of these
  1. Which satellite used to view polar and equatorial regions?
  1. Write note on polar satellites
  1. Write a note on geostationary satellite. Give its applications.
Answer
  1. (a) 24 hours
  1. 500 to 800km
  1. Polar satellites are used to view polar and equatorial regions as they rotate on poles of earth.
  1. Polar satellites are low altitude (500 to 800 km) satellites, but they go around the poles of the earth in a north-south direction. Since its time period is around 100 minutes it crosses any altitude many times a day. Information gathered from such satellites is extremely useful for remote sensing, meterology as well as for environmental studies of the earth.
  1. Satellites in circular orbits around the earth in the equatorial plane with time period same as earth are called Geostationary Satellites.
Applications:- Radio waves broadcast. Satellites widely used for telecommunications in India. GPS system, navigation system , defence etc.
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Question 74 Marks
Read the passage given below and answer the following questions from 1 to 5. Satellites in a circular orbits around the earth in the equatorial plane with T = 24 hours are called Geostationary Satellites. Clearly, since the earth rotates with the same period, the satellite would appear fixed from any point on earth. It takes very powerful rockets to throw up a satellite to such large heights above the earth but this has been done in view of the several benefits of much practical application. Weight of an object is the force with which the earth attracts it. We are conscious of our own weight when we stand on a surface, since the surface exerts a force opposite to our weight to keep us at rest. The same principle holds good when we measure the weight of an object by a Spring balance hung from a fixed point e.g. the ceiling. The object would fall down unless it is subject to a force opposite to gravity. This is exactly what the spring exerts on the object. This is because the spring is pulled down a little by the gravitational pull of the object and in turn the spring exerts a force on the object vertically upwards. Now, imagine that the top end of the balance is no longer held fixed to the top ceiling of the room. Both ends of the spring as well as the object move with identical acceleration g. The spring is not stretched and does not exert any upward force on the object which is moving down with acceleration g due to gravity. The reading recorded in the spring balance is zero since the spring is not stretched at all. If the object were a human being, he or she will not feel his weight since there is no upward force on him. Thus, when an object is in free fall, it is weightless and this phenomenon is usually called the phenomenon of weightlessness. In a satellite around the earth, every part and parcel of the satellite has acceleration towards the centre of the earth which is exactly the value of earth’s acceleration due to gravity at that position. Thus in the satellite everything inside it is in a state of free fall. This is just as if we were falling towards the earth from a height. Thus, in a manned satellite, people inside experience no gravity. Gravity for us defines the vertical direction and thus for them there are no horizontal or vertical directions, all directions are the same.
  1. Astronaut experiences weightlessness in space because:
  1. Acceleration due to gravity is zero.
  2. Actual weight of astronaut is zero.
  3. They are going with same acceleration due to gravity.
  4. None of these.
  1. Weighing machine measures:
  1. Mass of the person.
  2. Normal reaction exerted by machine on person.
  3. Both a and b.
  4. None of these.
  1. What is geostationary satellite?
  1. What is weight? How it is measured?
  1. What is weightlessness astronaut in satellite experienced by ?
Answer
  1. (c) They are going with same acceleration due to gravity.
  1. Normal reaction exerted by machine on person.
  1. Satellites in a circular orbits around the earth in the equatorial plane with T = 24 hours are called Geostationary Satellites. Clearly, since the earth rotates with the same period, the satellite looks like stationary object from earth.
  1. Weight of an object is the force with which the earth attracts it. It is measured with the help of spring balance.
  1. Weightlessness is condition in which acceleration due to gravity is balanced by satellite as it is moving and astronaut don’t feel any weight hence called weightlessness. In a satellite around the earth, every part and parcel of the satellite has acceleration towards the centre of the earth which is exactly the value of earth’s acceleration due to gravity at that position. Thus in the satellite everything inside it is in a state of free fall. Thus, in a manned satellite, people inside experience no gravity.
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Question 84 Marks
Read the passage given below and answer the following questions from 1 to 5. Earth’s Satellite: Earth satellites are objects which revolve around the earth. Their motion is very similar to the motion of planets around the Sun. In particular, their orbits around the earth are circular or elliptic. Moon is the only natural satellite of the earth with a near circular orbit with a time period of approximately $27.3$ days which is also roughly equal to the rotational period of the moon about its own axis. Also, the speed that a satellite needs to be travelling to break free of a planet or moon’s gravity well and leave it without further propulsion is known as escape velocity. For example, a spacecraft leaving the surface of earth needs to be going 7 miles per second or nearly 25000 miles per hour to leave without falling back to the surface or falling into orbit.
  1. Gas escapes from the surface of a planet because it acquires an escape velocity. The escape velocity will depend on which of the following factors?
  1. Mass of the planet
  2. Mass of the particle escaping
  3. Temperature of the planet
  4. None of the above
  1. The escape velocity of a satellite from the earth is ve If the radius of earth contracts to $(\frac{1}{4})$ th of its value, keeping the mass of the earth constant, escape velocity will be:
  1. doubled
  2. halved
  3. tripled
  4. unaltered
  1. The ratio of escape velocity at earth $(v_e)$ to the escape velocity at a planet $(v_p)$, whose radius and mean density are twice as that of earth is:
  1. $1:2\sqrt{2}$
  2. 1 : 4
  3. $1:\sqrt{2}$
  4. 1 : 2
  1. A satellite S is moving in an elliptical orbit around the earth. The mass of the satellite is very small as compared to the mass of the earth, then:
  1. the angular momentum of S about the centre of the earth changes in direction, but its magnitude remains constant
  2. the total mechanical energy of S varies periodically with time
  3. the linear momentum of S remains constant in magnitude
  4. the acceleration of S is always directed towards the centre of the earth
  1. The orbital velocity of an artificial satellite in a circular orbit just above the earth’s surface is vo The orbital velocity of a satellite orbiting at an altitude of half of the radius, is:
  1. $\frac{3}{2}\text{v}_\circ$
  2. $\frac{2}{3}\text{v}_\circ$
  3. $\sqrt{\frac{3}{2}\text{v}_\circ}$
  4. $\sqrt\frac{2}{3}\text{v}_\circ$
Answer
  1. (a) Mass of the planet
Explanation:
As we know that, escape velocity,
$\text{v}_\text{e}=\sqrt\frac{2\text{Gm}}{\text{R}}$
where, M is mass of planet. So, on the basis of Eq. (i), it can be said that escape velocity will depend upon the mass of the planet (M).
  1. (a) doubled
Explanation:
Given, escape velocity on the surface of earth,
$\text{v}_\text{e}=\sqrt\frac{2\text{Gm}_\text{e}}{\text{R}_\text{e}}$
where, $M_e$ = mass of the earth and $R_e$ = radius of the earth.
Now, according to the question, radius of earth,
$\acute{\text{R}}=\frac{\acute{\text{R}}}{4}$
$\Rightarrow\acute{\text{v}}_\text{e}=\sqrt\frac{2\text{GM}_\text{e}}{\text{R}_\text{e}}=\sqrt{4\big(\frac{2\text{GM}_\text{e}}{\text{R}_\text{e}}}\big)$
$=2\sqrt{\frac{2\text{GM}_\text{e}}{\text{R}_\text{e}}}$
or $\acute{\text{v}}_\text{e}=2\acute{\text{v}_\text{e}}$
  1. a $1:2\sqrt{2}$
Explanation:
Since, the escape velocity of earth can be given as
$\text{v}_\text{e}=\sqrt{2\text{gr}}$
$\Rightarrow\text{v}_\text{e}=\text{R}\sqrt{\frac{8}{3}\pi\text{G}\rho}$
($\rho=$ density of earth)...(i)
As it is given that, the radius and mean density of planet are twice as that of earth. So, escape velocity at planet will be
$\text{v}_\text{e}=\text{R}\sqrt{\frac{8}{3}\pi\text{G2}\rho}$...(ii)
Dividing Eq. (i) by Eq. (ii), we get
$\frac{\text{v}_\text{e}}{\text{v}_\rho}=\frac{{\text{R}\sqrt{\frac{8}{3}\pi\text{G}\rho}}}{2\text{R}\sqrt{\frac{8}{3}\pi\text{G}\rho}}$
$\Rightarrow\frac{\text{v}_\text{e}}{\text{v}_\text{e}}=\frac{1}{2\sqrt{2}}=1:2\sqrt{2}$
  1. (d) the acceleration of S is always directed towards the centre of the earth
Explanation:
As, we know that, force on satellite is only gravitational force which will always be towards the centre of earth. Thus, the acceleration of S is always directed towards the centre of the earth.
  1. d$\sqrt\frac{2}{3}\text{v}_\circ$
Explanation:
Orbital velocity is given by $\text{v}_\circ=\sqrt{\frac{\text{GM}}{\text{r}}}$
where, r = R + h
If $\text{h}=\frac{\text{R}}{2},$ Then $\text{r}=\text{R}+\frac{\text{R}}{2}=\frac{3}{2}\text{R}$
Then orbital velocity of satellite orbiting at half altitude becomes,
$\therefore\text{v}=\sqrt\frac{\text{GM}\times2}{3\text{R}}=\sqrt\frac{2}{3}\text{v}_\circ$
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