CBSE Test Paper 03

Chapter 8 Gravitation

1. A spaceship is stationed on Mars. How much energy must be expended on the

spaceship to launch it out of the solar system? Mass of the space ship = 1000 kg; mass
of the sun = 2 kg; mass of mars = 6.4 kg; radius of mars = 3395 km;

radius of the orbit of mars = 2.28 km; G = 6.67 N m2 kg .1

a. 5.5 J
b. 5.66 J
c. 5.96 J
d. 5.86 J

2. For a satellite to be in a circular orbit 780 km above the surface of the earth, what is
the period of the orbit (in hours)? 1

a. 1.98 hr
b. 1.65 hr
c. 1.78 hr
d. 1.88 hr

3. A body of mass m is taken from earth surface to the height h equal to radius of earth,
the increase in potential energy will be 1

a.
b. mgR
c.
d. 2mgR

4. Titania, the largest moon of the planet Uranus, has the radius of the earth and
the mass of the earth. What is the average density of Titania? Data: G = 6.67 N

m2/kg2, RE = 6.38 m, = 5.97 kg. 1

a. 2300 kg/m3

b. 1900 kg/m3

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 c. 2700 kg/m3

d. 1700 kg/m3

5. If the radius of earth reduces by 4% and density remains same then escape velocity
will 1

a. Increase by 2%
b. Reduce by 4%
c. Increase by 4%
d. Reduce by 2%

6. Two satellites are at different heights. Which would have greater velocity? 1

7. By which law is the Kepler’s law of areas identical? 1

8. Where does a body weight more; at the surface of the earth or in the mine? 1

9. A uniform ring of mass m and radius a is placed directly above a uniform sphere of
mass M and of equal radius. The centre of the ring is at a distance 33 a from the centre
of the sphere. Find the gravitational force exerted by the sphere on the ring. 2

10. State and briefly explain the law of areas. 2

11. A body hanging from a spring stretches it by 1 cm at the earth's surface. How much
will the same body stretch at a place 1600 klm above the earth's surface? Radius of
earth 6400 km. 2

12. Let the speed of the planet at perihelion P in fig be vp and Sun planet distance SP be

rp. Relate (rp, vp) to the corresponding quantities at the aphelion (rA, vA). Will the

planet take equal times to traverse BAC and CPB? 3

13. A satellite orbits the earth at a height of 400 km above the surface. How much energy
must be expended to rocket the satellite out of the earth's gravitational influence?

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 Mass of the satellite = 200 kg; mass of the earth = kg; radius of the earth =
m; G = .3

14. A star 2.5 times the mass of the sun and collapsed to a size of 12 km rotates with a
speed of 1.2 rev. per second. (Extremely compact stars of this kind are known as
neutron stars. Certain stellar objects called pulsars belong to this category). Will an
object placed on its equator remain stuck to its surface due to gravity? (Mass of the

sun = 2 1030 kg). 3

15. Show the nature of the following graph for a satellite orbiting the earth. 5

i. K.E. versus orbital radius R.

ii. P.E. versus orbital radius R.
iii. T.E. versus orbital radius R.

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 CBSE Test Paper 03
Chapter 8 Gravitation

Answer

1. c. 5.96 J
Explanation: Mass of the spaceship, ms = 1000 kg

Mass of the Sun, M = 2 1030kg

Mass of Mars, Mm = 6.4 1023kg

Orbital radius of Mars, r = 3395 km = 3.395 106m

Universal gravitational constant, G = 6.67 10-11 Nm2kg-2

Potential energy of the spaceship due to gravitational attraction of the Sun = -

Potential energy of the spaceship due to gravitational attraction of the Mars = -

Since the spaceship is stationed on Mars, its velocity and hence, its kinetic
energy will be zero.
Total energy of the spaceship =

The negative sign indicates that the system is in bound state.

Energy required for launching the spaceship out of the solar system

= - Total energy of the spaceship =

= 6.67 10-8 (87.72 1017 + 1.88 1017)

= 6.67 10-8 89.50 1017

= 596.96 109

= 5.96 1011 J

2. b. 1.65 hr

Explanation: Mass of the Earth, Me = 6.0 1024kg

Radius of the Earth,

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 Universal gravitational constan t, G =

Height of the satellite, h = 780 km = 780 103m = 0.78 106m

Time Period of the satellite, T =

= 5958.85 sec = 1.65 hr

3. a.
Explanation: Work done

If h = R then,

4. d. 1700 kg/m3

Explanation: Density of Titania =

Here, mass of earth (ME = 5.97 1024kg) and radius of earth(RE = 6.38 106m)

= 1.7 103kg/m3 = 1700 kg/m3

5. b. Reduce by 4%
Explanation: Escape velocity ve R and if density remains constant

R
So if the radius reduces by 4% then escape velocity also reduces by 4%

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 6. va . The velocity is inversely proportional to the height. Therefore the satellite at

smaller height would possess greater velocity.

7. The Kepler’s law of areas is identical to the law of conservation of angular

momentum.

8. The value of g in mine is less than that on the surface of the earth. Therefore weight
will be more on the surface of earth as compared to the mines.

9. The gravitational field at any point on the ring due to the sphere is equal to the field
due to a single particle of mass M placed at the centre of the sphere. Thus, the force on
the ring due to the sphere is also equal to the force on it by a particle of mass M placed
at this point. By Newton's third law it is equal to the force on the particle by the ring.
Now the gravitational field due to the ring at a distance d = a on its axis is

E= =

The force on a particle of mass M placed here is F = ME

=
This is also the force due to the sphere on the ring.

10.

The law of areas or Kepler's second law of planetary motion states that the line
joining Sun and the planet sweeps out equal areas in equal intervals of time,
howsoever small these intervals may be. Alternately, we can say that the areal
velocity of a planet remains constant.

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 In Figure, if a planet moves from P to P' in a small time interval At, then area swept
A = area SPP'

and = a constant quantity

11. In equilibrium mg = kg,

At height h mg’ = kx’,

12. The magnitude of angular momentum at P, Lp = mprpvp

Similarly, magnitude of angular momentum at P is LA = mArAvA

From conservation of angular momentum

mprpvp = mArAvA (mp = mA)

as rA > rp vp > vA

The magnitude of angular momentum at P, Lp = mprpvp

Similarly, magnitude of angular momentum at A is LA = mArAvA

From conservation of angular momentum

mprpvp = mArAvA (mp = mA)

as
area bounded by SB & SC are (SBAC > SBPC)
By 2nd law equal areas are swept in equal intervals of time. Time taken to
transverse BAC > time taken to traverse CPB.

13.
Mass of the satellite, m = 200 kg

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 Height of the satellite, h = 400 km
Total energy of the satellite at height h =Kinectic energy + Potential energy

Total energy of the satellite at height h

Orbital velocity of the satellite,

Total energy=

Energy required to send the satellite out of its orbit = -(Bound energy)

14. For the object to remain stuck to the star, the gravitational force of the star must be
equal to or greater than the centripetal force. Under this condition, the centrifugal
force does not overcome the gravitational force and not fly off the object. It means

mg > mv2/r

g > v2/r
g > ac

Where ac=v2/r

Centripital acceleration
for the object to remain stuck, the acceleration due to gravity (g) on the star must be >
centripetal acceleration
Gravitational force,

Where,
M = Mass of the star =
m = Mass of the body
R = Radius of the star = 12 km = 1.2 ×104 m

Centrifugal force,
= Angular speed = 2πv

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 v = Angular frequency = 1.2 rev s-1

Since , the body will remain stuck to the surface of the star.

15. Mass of earth

Radius of orbit of satellite= R
Mass of satellite = m

Orbital Velocity

a. EK versus R:

we can say that kinetic energy is inversely proportional to R. It means

the KE decreases exponentially with radius. The graph will be a rectangular
hyperbola. Hence, the variation of kinetic energy versus orbital radius is shown in
graph. decreases exponentially with R.
b. Ep versus R: potential energy is twice of kinetic energy and negative sign implies

that graph is downward hyperbola.

c. T.E. versus R: Negative total energy, E signifies that earth and the satellite is a
bounded system.
If E ≥ U, the satellite will be free from earth’s gravity

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