Practice Start Test

Gravitation

NEET Physics · 96 questions · Page 3 of 10 · Click an option or "Show Solution" to reveal answer

Q21
The kinetic energies of a planet in an elliptical orbit about the Sun, at positions A, B and C are K A , K B and K C , respectively. AC is the major axis and SB is perpendicular to AC at the position of the Sun S as shown in the figure. Then
A K A < K B < K C
B K A > K B > K C
C K B < K A < K C
D K B > K A > K C
Correct Answer
Option B
Solution

Speed of the planet will be maximum when its distance from the sun is minimum as mvr = constant.

Point A is perihelion and C is aphelion.

Clearly, V A > V B > V C So, K A > K B > K C

Q22
The acceleration due to gravity at a height at a height 1 km above the rearth is the same as at a depth d below the surface of earth. Then
A d = 1 km
B d = 32{3 \over 2} km
C d = 2 km
D d = 12{1 \over 2} km
Correct Answer
Option C
Solution

Above earth surface

gh=g(1−2hRe){g_h} = g\left( {1 - {{2h} \over {{R_e}}}} \right)

Below earth surface

gd=g(1−dRe){g_d} = g\left( {1 - {d \over {{R_e}}}} \right)

According to question, g h = g d

g(1−2hRe)g\left( {1 - {{2h} \over {{R_e}}}} \right)

=

g(1−dRe)g\left( {1 - {d \over {{R_e}}}} \right)

Clearly, d = 2h = 2 km

Q23
Two astronauts are floating in gravitational free space after having lost contact with their spaceship. The two will
A move towards each other.
B move away from each other.
C will become stationary.
D keep floating at the same distance between them.
Correct Answer
Option A
Solution

Since two astronauts are floating in gravitational free space.

The only force acting on the two astronauts is the gravitational pull of their masses,

F=Gm1m2r2F = {{G{m_1}{m_2}} \over {{r^2}}}

which is attractive in nature. Hence they move towards each other.

Q24
A satellite of mass m is orbiting the earth (of radius R) at a height h from its surface. The total energy of the satellite in terms of g 0 , the value of acceleration due to gravity at the earth's surface, is
A mg0R22(R+h){{m{g_0}{R^2}} \over {2\left( {R + h} \right)}}
B −mg0R22(R+h) - {{m{g_0}{R^2}} \over {2\left( {R + h} \right)}}
C 2mg0R2R+h{{2m{g_0}{R^2}} \over {R + h}}
D −2mg0R2R+h - {{2m{g_0}{R^2}} \over {R + h}}
Correct Answer
Option B
Solution

Total mechanical energy of satellite is given by E = – GMm/2r Here, r = R + h And, GM = g o R 2 So, E = – mg o R 2 /2(R+h)

Q25
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
A 1 : 4
B 1 : 2\sqrt 2
C 1 : 2
D 1 : 22\sqrt 2
Correct Answer
Option D
Solution

As we know, escape velocity,

Ve=2GMR=2GR.(43πR3ρ)∝Rρ{V_e} = \sqrt {{{2GM} \over R}} = \sqrt {{{2G} \over R}.\left( {{4 \over 3}\pi {R^3}\rho } \right)} \propto R\sqrt \rho

∴\therefore

VeVp=ReRpρeρp{{{V_e}} \over {{V_p}}} = {{{R_e}} \over {{R_p}}}\sqrt {{{{\rho _e}} \over {{\rho _p}}}}
⇒VeVp=Re2Reρe2ρe\Rightarrow {{{V_e}} \over {{V_p}}} = {{{R_e}} \over {2{R_e}}}\sqrt {{{{\rho _e}} \over {2{\rho _e}}}}

∴\therefore Ration

VeVp=1:22{{{V_e}} \over {{V_p}}} = 1:2\sqrt 2
Q26
At what height from the surface of earth the gravitation potential and the value of g are −-5.4 × \times 10 7 J kg −-1 and 6.0 m s −-2 respectively? Take the radius of earth as 6400 km.
A 1400 km
B 2000 km
C 2600 km
D 1600 km
Correct Answer
Option C
Solution

As we know, gravitational potential (v) and acceleration due to gravity (g) with height

V=−GMR+h=−5.4×107V = {{ - GM} \over {R + h}} = - 5.4 \times {10^7}

...(i) and

g=GM(R+h)2=6g = {{GM} \over {{{\left( {R + h} \right)}^2}}} = 6

...(ii) Dividing (i) by (ii)

−GMR+hGM(R+h)2=−5.4×1076{{{{ - GM} \over {R + h}}} \over {{{GM} \over {{{\left( {R + h} \right)}^2}}}}} = {{ - 5.4 \times {{10}^7}} \over 6}
⇒−5.4×107(R+h)=6\Rightarrow {{ - 5.4 \times {{10}^7}} \over {\left( {R + h} \right)}} = 6

⇒\Rightarrow R + h = 9000 km so, h = 2600 km

Q27
A satellite S is moving in an elliptical orbit around the earth. The mass of the satellite is very small compared to the mass of the earth. Then,
A the linear momentum of S remains constant in magnitude.
B the acceleration of S is always directed towards the centre of the earth.
C the angular momentum of S about the centre of the earth changes in direction, but its magnitude remains constant.
D the total mechanical energy of S varies periodically with time.
Correct Answer
Option B
Solution

The gravitational force on the satellite will be aiming towards the centre of the earth so acceleration of the satellite will also be aiming towards the centre of the earth.

Q28
A remote-sensing satellite of earth revolves in a circular orbit at a height of 0.25 × \times 10 6 m above the surface of earth. If earth's radius is 6.38 × \times 10 6 m and g = 9.8 ms −-2 , then the orbital speed of the satellite is
A 9.13 km s −-1
B 6.67 km s −-1
C 7.76 km s −-1
D 8.56 km s −-1
Correct Answer
Option C
Solution

The orbital speed of the satellite is

vo=Rg(R+h){v_o} = R\sqrt {{g \over {\left( {R + h} \right)}}}

where R is the earth’s radius, g is the acceleration due to gravity on earth’s surface and h is the height above the surface of earth.

Here, R = 6.38 × 10 6 m, g = 9.8 m s –2 and h = 0.25 × 10 6 m

vo=(6.38×106m)9.8 ms−2(6.38×106m+0.25×106m){v_o} = \left( {6.38 \times {{10}^6}m} \right)\sqrt {{{9.8\,m{s^{ - 2}}} \over {\left( {6.38 \times {{10}^6}m + 0.25 \times {{10}^6}m} \right)}}}

= 7.76 × 10 3 m s –1 = 7.76 km s –1

Q29
Kepler's third law states that square of period of revoluation (T) of a planet around the sun, is proportional to third power of average distance r between sun and planet i.e. T 2 = Kr 3 here K is constant. If the masses of sun and planet are M and m respectively then as per Newton's law of gravitation force of attraction between them is F = GMmr2{{GMm} \over {{r^2}}}, here G is gravitational constant. The relation between G and K is described as
A K = G
B K = 1G{1 \over G}
C GK = 4π\pi 2
D GMK = 4π\pi 2
Correct Answer
Option D
Solution

Gravitational force of attraction between planet and sun gives centripetal force, GMm/r 2 = mv 2 /r Now velocity v =

GM/r\sqrt {GM/r}

Time period of planet T = 2π\pir/v ⇒\Rightarrow T 2 = 4

π2\pi ^2

r 3 /GM From Kepler’s third law, T 2 = Kr 3 Using equations, we see 4π\pi 2 r 3 /GM = Kr 3 , so relation between G and K is GMK = 4

π2\pi ^2
Q30
A black hole is an object whose gravitational field is so strong that even light cannot escape from it. To what approximate radius would earth (mass = 5.98 × \times 10 24 kg) have to be compressed to be a black hole?
A 10 −-9 m
B 10 −-6 m
C 10 −-2 m
D 100 m
Correct Answer
Option C
Solution

From question, Escape velocity

=2GMR=c= \sqrt {{{2GM} \over R}} = c

= speed of light

⇒R=2GMc2\Rightarrow R = {{2GM} \over {{c^2}}}
=2×6.6×10−11×5.98×1024(3×108)2m= {{2 \times 6.6 \times {{10}^{ - 11}} \times 5.98 \times {{10}^{24}}} \over {{{\left( {3 \times {{10}^8}} \right)}^2}}}m

= 10 -2 m

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