# 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}$$\frac{{{(R+h)}^{3/2}}}{{{g}^{1/2}}} Or, (R+h)3/2 = \frac{TR{{g}^{1/2}}}{2\pi } Or, R + h = {{\left( \frac{TR{{g}^{1/2}}}{2\pi } \right)}^{2/3}} Or, h = {{\left( \frac{TR{{g}^{1/2}}}{2\pi } \right)}^{2/3}}- R Or, h = {{\left( \frac{9000\times 6.37\times {{10}^{6}}\times {{10}^{1/2}}}{2\pi } \right)}^{2/3}}- 6.37×106 h = 3037.36 Km Method-2: We know, h = {{\left( \frac{{{T}^{2}}{{R}^{2}}g}{4{{\pi }^{2}}} \right)}^{1/3}}–R or, h = 3037.36 km \therefore The height of the satellite above the earth is 3037.36 km. ### 7. Obtain the value of g from the motion of the moon assuming that its period of rotation around the earth is 27 days 8 hours and the radius of the orbit is 60.1 times the radius of the earth. Solution, Here Acceleration due to gravity on earth, g = ? Time period, T = 27 days 8 hours = (27 × 24 + 8) × 60 × 60 = 2361600 Sec We know, radius of earth R = 6.4×106 m If the radius of orbit of satellite be ‘r’ then we can write, r = 60.1 R = 60.1 × 6.4×106 m We have the time period of satellite, T = \frac{2\pi }{R}$$\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

### 4 thoughts on “Gravitation Important Numerical Questions (solved) Class 11 Physics”

1. It is very useful for us sir good work sir

2. All these questions are formula based
Plz gave us conceptual question
By the way thank you sir

3. thank you so much !You helped lot of students like us !Keep going sir

4. thank you so much for this

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