Gravitation Important Numerical Questions (solved) Class 11 Physics

Formulae for Solving Gravitation Numerical Problems:

1. Newton’s law of Gravitation.

F = $\frac{G{{m}_{1}}{{m}_{2}}}{{{d}^{2}}}$, where G is universal gravitational constant.

The value of G is 6.67×10–11 Nm2kg–2

2. The expression for acceleration due to gravity on the surface of earth or large mass.

  g = $\frac{GM}{{{R}^{2}}}$

3. Variation of ‘g’ with height/altitude (h).

  g’ = g$\left( 1-\frac{2h}{R} \right)$, 

4. Variation of ‘g’ with depth ‘d’. 

g’ = g $\left( 1-\frac{d}{R} \right)$

5. Variation of ‘g’ with rotation of earth. / Variation of ‘g’ with latitude. 

g’ = g –${{\omega }^{2}}R{{\cos }^{2}}\theta $, 

6. Gravitation Field Intensity (I).

Then, I = $\frac{GM}{{{R}^{2}}}$

7. Gravitational Potential (V or vg)

Vg = $\frac{-GM}{R+h}$

8. Gravitational Potential Energy (U)

U = -$\frac{GMm}{r}$  

9. Escape Velocity (Ve),

Ve = $\sqrt{\frac{2GM}{R}}$

Ve = $\sqrt{2gR}$ ∴

Ve = 11.2 kms–1, which is the escape velocity of the earth.

For Satellite:

10. Gravitational force provides necessary centripetal force i.e.

$\frac{m{{V}_{o}}^{2}}{r}$= $\frac{GMm}{{{r}^{2}}}$

11. Orbital Velocity 

  V0 = R $\sqrt{\frac{g}{R+h}}$ 

Also, Vo = $\sqrt{\frac{GM}{r}}$ [r = R+h]

12. Time period of satellite,

T = $\frac{2\pi }{R}$$\sqrt{\frac{{{(R+h)}^{3}}}{g}}$ 

13. Height (h) of the satellite,

h = ${{\left( \frac{{{T}^{2}}{{R}^{2}}g}{4{{\pi }^{2}}} \right)}^{1/3}}$–R

14. Total Energy associated with a satellite.

Etotal = K.E. + G.P.E.

15. K.E. = $\frac{1}{2}$m Vo2 = $\frac{GMm}{2r}$

16. Gravitational Potential Energy (U)

U = -$\frac{GMm}{r}$  

17. Total Energy associated with a satellite.

Etotal = –$\frac{GMm}{2r}$= –$\frac{GMm}{2(R+h)}$

Numerical Problems of Gravitation:

1. Calculate the point along a line joining the centers of earth and moon where there is no gravitational force.  

Solution,
Gravitation Important Numerical Questions (solved) Class 11 Physics

Let X be the distance of the point from the earth along a line joining the centers of earth and moon where there is no gravitational force

Mass of earth, Me = 6×1024 kg

Mass of moon, Mm = 7.4×1022 kg

Distance between their, d = 3.8×108 m

For no gravitational force, if we place an object of mass ‘m’ then 

Gravitational force between the earth and the object = Gravitational force between the moon and the object. i.e. 

$\frac{G{{M}_{e}}m}{{{x}^{2}}}$= $\frac{G{{M}_{m}}m}{{{(d-x)}^{2}}}$

Or, $\frac{{{M}_{e}}}{{{M}_{m}}}$ = ${{\left( \frac{x}{d-x} \right)}^{2}}$

Or, $\frac{x}{d-x}$= $\sqrt{\frac{6\times {{10}^{24}}}{7.4\times {{10}^{22}}}}$= 9.0045

Or, x = 9.0045d –9.0045x

Or, x = $\frac{9.0045\times 3.8\times {{10}^{8}}}{1+9.0045}$

$\therefore $ x = 3.42×108 m from the earth.

2. Mass of earth is 81 times heavier than the moon. Its diameter is about 4 times larger than that of the moon. Estimate the value of acceleration due to gravity on the surface of the moon. 

Solution,

Let M and m be the masses, D and d be the diameter and R and r be the radius of earth and moon respectively.

M = 81m

D = 4d

Or, 2R = 4×2r

Or R = 4r

Acceleration due to gravity on the surface of moon, gm = ?

On the surface of earth, acceleration due to gravity

ge = $\frac{GM}{{{R}^{2}}}$……..(i)

On the surface of moon, acceleration due to gravity

gm = $\frac{Gm}{{{r}^{2}}}$……..(ii)

Dividing equation (ii) by (i),

$\frac{{{g}_{e}}}{{{g}_{m}}}$ = $\frac{m}{{{r}^{2}}}$× $\frac{{{R}^{2}}}{M}$

Or, gm = ge $\frac{m}{{{r}^{2}}}$× $\frac{{{(4r)}^{2}}}{M}$

Or, gm = ge $\frac{m}{{{r}^{2}}}$× $\frac{{{(4r)}^{2}}}{M}$

Or, gm = ge $\frac{m}{{{r}^{2}}}$× $\frac{{{(4r)}^{2}}}{81m}$

Or, gm = 9.8 × $\frac{16}{81}$= 1.94 ms–2

3. A man can jump 1.5m on earth. Calculate the approximate height he might be able to jump on a planet whose density is one quarter of the earth and where radius is one third that of the earth.

Solution,

Here,

Height on earth, hE = 1.5m

Height on the planet, hp = ?

If ρP and ρE be the density and  RP and RE be the radius of planet and earth respectively then according to question,

${{\rho }_{P}}=\frac{{{\rho }_{E}}}{4}$ & RP = $\frac{{{R}_{E}}}{3}$

For a given man, Initial K.E. of man at planet = Initial K.E. of man at earth.

and according to Work energy theorem, we can write 

Potential Energy at planet = Potential energy at earth 

Or, mgphp = mgEhE

Or, hP = hE $\frac{{{g}_{E}}}{{{g}_{P}}}$………. (i)

Since g = $\frac{GM}{{{R}^{2}}}$

Or, hP = hE $\frac{\frac{G{{M}_{E}}}{{{R}_{E}}^{2}}}{\frac{G{{M}_{P}}}{{{R}_{P}}^{2}}}$

Or, hP = hE ×$\frac{{{V}_{E}}{{\rho }_{E}}}{{{R}_{E}}^{2}}$× $\frac{R{{{}_{P}}^{2}}}{{{V}_{P}}{{\rho }_{P}}}$

Or, hP = hE ×$\frac{4/3\pi {{R}_{E}}^{3}\rho {}_{E}}{{{R}_{E}}^{2}}$× $\frac{{{R}_{P}}^{2}}{4/3\pi {{R}_{P}}^{3}{{\rho }_{P}}}$

Or, hP = hE× RE ${{\rho }_{E}}$× $\frac{3\times 4}{{{R}_{E}}{{\rho }_{E}}}$

Or, hP =1.5×3×4 = 18m

$\therefore $ The man can jump upto a height of 18m.

4. Find the height of the geostationary satellite above the earth assuming earth as a sphere of radius 6370 km.

Solution,

Here,

The time period for geostationary satellite,

T = 24 hrs = 24×60×60

= 86400 sec

Radius of earth, R = 6370km = 6.37×106m

Height of satellite, h = ?

We know, time period,

T = $\frac{2\pi }{R}$$\sqrt{\frac{{{(R+h)}^{3}}}{g}}$

T = $\frac{2\pi }{R}$$\frac{{{(R+h)}^{3/2}}}{{{g}^{1/2}}}$

Or, (R+h)3/2 = $\frac{TR{{g}^{1/2}}}{2\pi }$

Or, R + h = ${{\left( \frac{TR{{g}^{1/2}}}{2\pi } \right)}^{2/3}}$

Or, h = ${{\left( \frac{TR{{g}^{1/2}}}{2\pi } \right)}^{2/3}}$- R

 Or, h = ${{\left( \frac{86400\times 6.37\times {{10}^{6}}\times {{10}^{1/2}}}{2\pi } \right)}^{2/3}}$- 6.37×106

$\therefore $ h = 36122818.44 m = 36122.82 Km

5. A remote sensing satellite of the earth revolves in a circular orbit at a height of 250 km above the earth’s surface. What is the orbital speed and a period of revolution of the satellite?

Solution,

Here,

Height, h = 250 km = 250×103 km

Radius of earth, R = 6400 km = 6.4×106 m 

(i)  Orbital speed, V0 = ?

(ii) Time period, T = ?

We have

(i) Orbital speed, V0 = R $\sqrt{\frac{g}{R+h}}$ 

Or, V0 = 6.4×106  $\sqrt{\frac{10}{6.4\times {{10}^{6}}+250\times {{10}^{3}}}}$ 

Or, V0 = 7848.18 ms–1  

(ii) Time period

We have, time period,

T = $\frac{2\pi }{R}$$\sqrt{\frac{{{(R+h)}^{3}}}{g}}$ 

Or, T = $\frac{2\pi }{6.4\times {{10}^{6}}}$$\sqrt{\frac{{{(6.4\times {{10}^{6}}+250\times {{10}^{3}})}^{3}}}{10}}$ 

$\therefore $ T = 5324 sec = $\frac{5324}{60}$= 88.73 min

6. An artificial satellite revolves around the earth in 2.5 hours in a particular orbit. Find the height of the satellite above the earth assuming earth as a sphere of radius 6370 km.

Solution,

Here,

Time period, T = 2.5 hrs = 2.5×60×60= 9000 sec

Radius of earth, R = 6370km = 6.37×106m

Height of the satellite, h = ?

Method-1:

We know, time period

T = $\frac{2\pi }{R}$$\sqrt{\frac{{{(R+h)}^{3}}}{g}}$

T = $\frac{2\pi }{R}$$\frac{{{(R+h)}^{3/2}}}{{{g}^{1/2}}}$

Or, (R+h)3/2 = $\frac{TR{{g}^{1/2}}}{2\pi }$

Or, R + h = ${{\left( \frac{TR{{g}^{1/2}}}{2\pi } \right)}^{2/3}}$

Or, h = ${{\left( \frac{TR{{g}^{1/2}}}{2\pi } \right)}^{2/3}}$- R 

Or, h = ${{\left( \frac{9000\times 6.37\times {{10}^{6}}\times {{10}^{1/2}}}{2\pi } \right)}^{2/3}}$- 6.37×106

h =  3037.36 Km

Method-2:

We know,

h = ${{\left( \frac{{{T}^{2}}{{R}^{2}}g}{4{{\pi }^{2}}} \right)}^{1/3}}$–R

or, h = 3037.36 km  

$\therefore $ The height of the satellite above the earth is 3037.36 km.

7. Obtain the value of g from the motion of the moon assuming that its period of rotation around the earth is 27 days 8 hours and the radius of the orbit is 60.1 times the radius of the earth.

Solution,

Here

Acceleration due to gravity on earth, g = ? 

Time period, T = 27 days 8 hours 

= (27 × 24 + 8) × 60 × 60 

= 2361600 Sec

We know, radius of earth R = 6.4×106 m

If the radius of orbit of satellite be ‘r’ then we can write, 

r = 60.1 R = 60.1 × 6.4×106 m

We have the time period of satellite,

T = $\frac{2\pi }{R}$$\sqrt{\frac{{{(R+h)}^{3}}}{g}}$

Or, g = ${{\left( \frac{2\pi }{RT} \right)}^{2}}$r3 [since r = R + h]

Or, g = ${{\left( \frac{2\pi }{6.4\times {{10}^{6}}\times 2361600} \right)}^{2}}$(60.1 × 6.4×106) 3

$\therefore $ g = 9.83 ms–2

8. The period of moon revolving under the gravitational force of the earth is 27.3 days. Find the distance of the moon from the centre of the earth if the mass of earth is 5.97×1024kg. 

Solution,

Here,

Time period, T = 27.3 days = 27.3×24×60×60 

= 2358720 sec

The distance of the moon from the centre of the earth, r = ?

Mass of the earth, M = 5.97×1024 kg

We have the time period of satellite,

T = $\frac{2}{R}$$\sqrt{\frac{{{(R+h)}^{3}}}{g}}$ here, R + h = r

Or, T = $\frac{2\pi }{R}$$\sqrt{\frac{{{r}^{3}}}{g}}$

 Or, T = 2$\pi $$\sqrt{\frac{{{r}^{3}}}{g{{R}^{2}}}}$

Or, T = 2$\pi $$\sqrt{\frac{{{r}^{3}}}{GM}}$ [Since g = $\frac{GM}{{{R}^{2}}}$]

Or, r = ${{\left( \frac{{{T}^{2}}}{4{{\pi }^{2}}}\times GM \right)}^{1/3}}$

Or, r = ${{\left( \frac{23587202}{4{{\pi }^{2}}}\times 6.67\times {{10}^{-11}}\times 5.97\times {{10}^{24}} \right)}^{1/3}}$

$\therefore $ r = 3.828×108 m

9. An earth satellite moves in a circular orbit with a speed of 6.2 kms–1 . Find the time of one revolution and its centripetal acceleration. 

Solution,

Here,

Orbital velocity, Vo = 6.2 kms–1 = 6.2×103ms–1

(i) Time Period, T = ?

(ii) Centripetal acceleration, a= ?

We have, time period,

T = $\frac{2\pi }{R}$$\sqrt{\frac{{{(R+h)}^{3}}}{g}}$ …….(i)

Also,

(i) Orbital speed, V0 = R $\sqrt{\frac{g}{R+h}}$

Or, R + h = ${{\left( \frac{R}{{{V}_{o}}} \right)}^{2}}$×g 

Or, R + h = ${{\left( \frac{6.4\times {{10}^{6}}}{6.2\times {{10}^{3}}} \right)}^{2}}$×10 

Or, R + h = 10655567.12 m

From equation (i)

T = $\frac{2\pi }{6.4\times {{10}^{6}}}$$\sqrt{\frac{{{(10655567.12)}^{3}}}{10}}$ 

Or, T = 10798.5 sec = $\frac{10798.5}{60\times 60}$

$\therefore $ T = 2.9996 hrs = 3 hrs.

(ii) Centripetal acceleration,

ac = $\frac{{{V}^{2}}}{r}$ [ as Fc = $\frac{m{{V}^{2}}}{r}$]

Or, ac = $\frac{{{(6.2\times {{10}^{3}})}^{2}}}{10655567.12}$= 3.61 ms–2

10. What is the period of revolution of a satellite of mass m that orbits the earth in a circular path of radius 7880 km about 1500 km above the surface of the earth?

Solution,

Here,

Time period, T = ?

Radius of orbit, r = 7880 km = 7880×103 km

Height, h = 1500 km = km

We have,

r = R + h

Radius of earth, R = r –h 

= 7880×103 – 1500×103 

= 6.38×106 m 

Time period, T = ?

We have, time period

T = $\frac{2\pi }{R}$$\sqrt{\frac{{{(R+h)}^{3}}}{g}}$ 

Or, T = $\frac{2\pi }{6.38\times {{10}^{6}}}$$\sqrt{\frac{{{(6.38\times {{10}^{6}}+1500\times {{10}^{3}})}^{3}}}{10}}$ 

Or, T = 6888.88 sec = $\frac{6888.88}{60}$min

$\therefore $ T = 114.81 min = 1.91 hrs

11. Calculate the period of revolution of a satellite revolving at a distance of 20 km above the surface of the earth, (radius of the earth = 6400 km. (acceleration due to gravity due to earth = 10ms–2)

Solution,

Here,

Height of the satellite, h = 20 km = 20×103 km

Radius of earth, R = 6400 km = 6.4×106 m 

Time period, T = ?

We have, time period

T = $\frac{2\pi }{R}$$\sqrt{\frac{{{(R+h)}^{3}}}{g}}$ 

Or, T = $\frac{2\pi }{6.4\times {{10}^{6}}}$$\sqrt{\frac{{{(6.4\times {{10}^{6}}+20\times {{10}^{3}})}^{3}}}{10}}$ 

Or,          T = 5050.13 sec  

$\therefore $ T = $\frac{5050.13}{60}$= 84.17 min

12. Taking the earth to be a uniform sphere of radius 6400 km, calculate the total energy needed to raise a satellite of mass 1000 kg to a height of 600 km above the ground and set it into circular orbit at that altitude.

Solution,

Here,

Radius of earth, R = 6400 km = 6.4×106 m

Total energy needed to launch the satellite, T.E = ?

Mass of satellite, m = 1000 kg

Height of the satellite, h = 600 km = 600×103 m

The total energy needed to raise a satellite to the given height and set it into circular orbit,

Etotal = increase in P.E. + K.E.

Or, Etotal = increase in P.E. + K.E.

Or, Etotal = (G.P.E.)r –(G.P.E.)R  + K.E.

Or, Etotal = $\frac{-GMm}{r}$ – $\left( \frac{-GMm}{R} \right)$ + $\frac{GMm}{2r}$

Or, Etotal = GMm $\left( \frac{-1}{r}+\frac{1}{R}+\frac{1}{2r} \right)$

Or, Etotal = gR2m $\left( \frac{-1}{r}+\frac{1}{R}+\frac{1}{2r} \right)$ [Since g = $\frac{GM}{{{R}^{2}}}$]

Or, Etotal = 10×(6.4×106)2×1000 $\left( \frac{-1}{6.4\times {{10}^{6}}+600\times {{10}^{3}}}+\frac{1}{6.4\times {{10}^{6}}}+\frac{1}{2(6.4\times {{10}^{6}}+600\times {{10}^{3}})} \right)$

$\therefore $ Etotal = 3.47×1010     J

13. Taking the earth to be a uniform sphere of radius 6400 km and the value of ‘g’ at the surface to be 10 m/s2, calculate the total energy needed to raise a satellite of mass 2000 kg to a height of 800 km above the ground and set it into circular orbit at that altitude.

Solution,

Here,

Radius of earth, R = 6400 km = 6.4×106

Acceleration due to gravity on earth, g = 10 ms–2

Total energy needed to launch the satellite, T.E = ?

Mass of satellite, m = 2000 kg

Height of the satellite, h = 800 km = 800×103 m

The total energy needed to raise a satellite to the given height and set it into circular orbit,

Etotal = increase in P.E. + K.E.

Or, Etotal = increase in P.E. + K.E.

Or, Etotal = (G.P.E.)r –(G.P.E.)R  + K.E.

Or, Etotal = $\frac{-GMm}{r}$ – $\left( \frac{-GMm}{R} \right)$ + $\frac{GMm}{2r}$

Or, Etotal = GMm $\left( \frac{-1}{r}+\frac{1}{R}+\frac{1}{2r} \right)$

Or, Etotal = gR2m $\left( \frac{-1}{r}+\frac{1}{R}+\frac{1}{2r} \right)$ [Since g = $\frac{GM}{{{R}^{2}}}$]

Or, Etotal = 10×(6.4×106)2×2000 $\left( \frac{-1}{6.4\times {{10}^{6}}+800\times {{10}^{3}}}+\frac{1}{6.4\times {{10}^{6}}}+\frac{1}{2(6.4\times {{10}^{6}}+800\times {{10}^{3}})} \right)$

$\therefore $ Etotal = 7.11×1010     J

14. A 200 kg satellite is lifted to an orbit of 2.2×104 km radius. If the radius and mass of the earth are 6.37×106 m and 5.98×1024 kg respectively. How much additional potential energy is required to lift the satellite?

Solution,

Here,

Mass of satellite, m = 200 kg

Radius of orbit, r = 2.2×104 km = 2.2×107 m

Radius of earth, R = 6.37×106

Mass of earth, M = 5.98×1024 kg 

Additional P.E. needed to raise the satellite to the given height = ?

Or, Additional P.E. = increase in G.P.E.

Or, Additional P.E. = increase in P.E.

Or, Additional P.E. = (G.P.E.)r –(G.P.E.)R  

Or, Additional P.E. = $\frac{-GMm}{r}$ – $\left( \frac{-GMm}{R} \right)$ 

Or, Additional P.E. = GMm $\left( \frac{-1}{r}+\frac{1}{R} \right)$

Or, Additional P.E. = 6.67×10–11 ×5.98×1024 ×200 $\left( \frac{-1}{2.2\times {{10}^{7}}}+\frac{1}{6.37\times {{10}^{6}}} \right)$

$\therefore $ Additional P.E. = 8.9×109   J

Also Read: Gravitation Notes Class 11

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