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#### Example – 01:

- Show that the escape velocity of a body orbiting very close to the earth’s surface of the earth is √2v
_{c}. Where v_{c}is the critical velocity of a body when it is orbiting very close to the earth’s surface. **Solution:**

For an orbiting satellite, critical velocity is given by

where G = Universal gravitational constant

M = the mass of the earth

R = the radius of the earth

h = height of the satellite above the earth’s surface.

- For satellite orbiting very close to the earth, h can be neglected as h < < R. Hence it can be neglected. Therefore the critical velocity of the satellite orbiting very close to earth’s surface.

The escape velocity of a satellite oh the surface of the earth is given by

Dividing equation (1) by (2)

#### Example – 02:

- Show that the escape velocity of a body from the surface of a planet of radius R and mean density ρ is

**Solution:**

The escape velocity of a satellite on the surface of the planet is given by

Where G = Universal gravitational constant

M = the mass of the Planet

R = the radius of the Planet

Let d = density of the material of the planet

Now, Mass of Planet = Volume of Planet x Density

This is an expression for escape velocity in terms of density of the material of the planet.

#### Example – 03:

- An artificial satellite has to be set up to revolve around the planet in a circular orbit close to its surface. If r is density and R is the radius of the planet, show that the period of revolution is

**Solution:**

The time period of a satellite orbiting around the earth is given by

If satellite orbiting very close to the earth (i.e. h < < R)

then h can be neglected. Then R + h = R

This is an expression for the period of a satellite in terms of density of the material of the planet.

#### Example – 04:

- Find the acceleration due to gravity on the surface of the moon. Given that the mass of the moon is 1/80 times that of the earth and the diameter of the moon is 1/4 times that of the earth. g = 9.8 m/s
^{2}. **Given:**M_{M}= M_{E}, R_{M}= R_{E}, g_{E}= 9.8 m/s^{2},**To find:**g_{M}= ?**Solution:**

Ans:The acceleration due to gravity on the surface of the moon is 1.96 m/s^{2}.

#### Example – 05:

- At what height above the earth’s surface will the acceleration due to gravity be 25% of the value at the surface of the earth? g= 9.8 m/s
^{2}, R= 6400 km. **Given:**R =6400 km, g_{h}= 25% g = 0.25 g**To find:**h =?**Solution:**

**Ans:** At height of 6400 km, the acceleration due to gravity be 25% of the value at the surface of the earth

#### Example – 06:

- Find the acceleration due to gravity at a depth of 2000 km from the surface of the earth, assuming earth to be a homogeneous sphere. R = 6400 km, g = 9.8 m/s
^{2}. **Solution:****Given: d**= 2000 km, Radius of earth = R = 6400 km, g = 9.8 m/s^{2}.**To find:**acceleration due to gravity = g_{d}=?

**Ans: **The acceleration at depth of 2000 km below the surface of the earth is 6,738 m/s^{2}

#### Example – 07:

- What would be the duration of the year if the distance of the Earth from the sun gets tripled?
**Given:**r_{1}= R, r_{2}= 32R , T_{1}= 1 year,**To find:**T_{2}=?**Solution:**

By Keppler’s law

**Ans:** The duration of the year will be 5.196 years.

#### Example – 08:

- Calculate the period of revolution of the planet Jupiter around the Sun. The ratio of the radius of Jupiter’s orbit to that of earth’s orbit around the Sun is 5.2.
**Solution:****Given:**r_{J}: r_{e }= 5.2 , Time period of the Earth T_{1}= 1 year.**To Find:**Period of Jupiter T_{J}=?

By Keppler’s law, we have T^{2} ∝ r^{3}

**Ans:** The period of revolution of planet Jupiter is 11.86 years.

#### Example – 09:

- Calculate the escape velocity of a body from the surface of the planet of diameter 2200 km. Acceleration due to gravity on the surface of the planet is 160 cm/s
^{2}. - Given diameter of planet = 2200 km, radius of planet = 1100 km= 1.1 x 10
^{6}m, Acceleration due to gravity = g = 160 cm/s^{2}= 1.6 m/s^{2}. - To Find: Escape velocity = Ve =?

**Ans:** The escape velocity on the surface of the planet is 1.876 km/s

#### Example – 10:

- What are the (1) KE (2) PE (3) total energy and (4) binding energy of an artificial satellite of mass 1000 kg orbiting at a height of 3600 km above the surface of the earth? Mass of the earth = 6 x 10
^{24}kg; Radius of earth = 6400 km, G = 6.67 × 10^{-11 }N m^{2}/kg^{2}, **Given:**Mass of satellite = m = 1000 kg, G = 6.67 × 10^{-11 }N m^{2}/kg^{2}, radius of earth = R = 6400 km; M = 6 x 10^{24}kg. h = 3600 km, r = R + h = 6400 + 3600 = 10000 km = 10 x 10^{6}m;**To Find:**K.E. = ?, P.E. = ?, T.E. = ?, B.E. = ?,**Solution:**

Now Total energy = – B.E. = – 2.0 x 10^{9} J

Now Potential Energy = – 2 x B.E. = – 2 x 2.0 x 10^{9} J = 4.0 x 10^{9} J

Kinetic energy = B.E. = 2.0 x 10^{9} J

**Ans: **The kinetic energy of satellite = 2.0 x 10^{9} J

Potential energy of satellite = – 4.0 x 10^{9} J

Total energy of satellite = – 2.0 x 10^{9} J

The binding energy of satellite = 2.0 x 10^{9} J

#### Example – 11:

- A body weights 4.5 kg wt on the surface of the earth. How much will it weigh on the surface of a planet whose mass is 1/9 that of earth and whose radius is half that of earth
**Solution****:****Given:**, Weight of body on earth = W_{E}= 4.5 kg wt, mass of planet = 1/9 mass of earth i.e M_{P}= 1/9 M_{E}, radius of planet = 1/2 radius of earth i.e. R_{P}= 1/2 R_{E}.**To find:**Weight of body on the planet = W_{P}= ?

**Ans: **The weight of the body on the surface of the planet is 2 kg wt.

#### Example – 12:

- The mass of mars is 1/9 that of earth and the radius is 1/2 that of earth. What is the weight of the body having mass of 1000 kg on earth.
**Solution****:****Given:**, mass of body on earth = m_{E}=1000 kg, mass of mars = 1/9 mass of earth i.e M_{P}= 1/9 M_{E}, radius of mars = 1/2 radius of earth i.e. R_{P}= 1/2 R_{E}.**To find:**Weight of body on the mars = W_{P}= ?

Weight on earth = 1000 x 9.8 = 9800 N

Mass of the body remains constant

**Ans: **The weight of the body on the surface of the mars is 4356 N

#### Example – 13:

- What is the decrease in the weight of a body of mass 500 kg when it is taken into a mine of depth 1000 m? R = 6400 km, g = 9.8 m/s
^{2}. **Solution:****Given:**Mass of body = m = 500 kg, depth d = 1000 m = 1 km, Radius of earth = R = 6400 km, g = 9.8 m/s^{2}.**To find:**decrease in weight = W – W_{d}=?

**Ans: **The decrease in the weight of the body is 1 N.

#### Example – 14:

- Find the weight of a body of mass 100 kg on the earth at a) equator b) pole c) latitude of 30°. R = 6400 km, g = 9.8 m/s
^{2} **Solution:****Given:**m =^{6}m,**To find:**W_{E}= ?, W_{P}= ? W_{Φ}= ?

Time period of earth = 24 hours = 24 x 60 x 60 s

The acceleration due to gravity at latitude Φ is given by

g_{Φ} = g – Rω^{2}cos^{2}Φ

The Weight of body at latitude Φ is given by

W_{Φ} = mg_{Φ} = mg – mRω^{2}cos^{2}Φ

At equator Φ = 0°

W_{E} = 100 x 9.8 – 100 x 6.4 × 10^{6 }x (7.273 x 10^{-5})^{2}cos^{2}0

W_{E} = 980 – 100 x 6.4 × 10^{6 }x (7.273 x 10^{-5})^{2 }x (1)^{2}

W_{E} = 980 – 3.386 = 976.6 N

At poles Φ =90°

W_{E} = 100 x 9.8 – 100 x 6.4 × 10^{6 }x (7.273 x 10^{-5})^{2}cos^{2}90

W_{E} = 980 – 100 x 6.4 × 10^{6 }x (7.273 x 10^{-5})^{2 }x (0)^{2}

W_{E} = 980 – 0 = 980 N

At latitude Φ = 30°

W_{E} = 100 x 9.8 – 100 x 6.4 × 10^{6 }x (7.273 x 10^{-5})^{2}cos^{2}30

W_{E} = 980 – 100 x 6.4 × 10^{6 }x (7.273 x 10^{-5})^{2 }x (0.866)^{2}

W_{E} = 980 – 2.539 = 977.5 N

Ans: Weight of the body on equator, on ploe and on latitude 30° are

976.6 N, 980 N and 977.5 N respectively.

#### Example – 15:

- Taking G = 6.67 × 10
^{-11 }N m^{2}/kg^{2}, the radius of the earth as 6400 km and mean density of earth as 5478 kg/m^{3}, calculate g at the surface of the earth. **Solution:****Given:**Radius of the Earth = R = 6400 km = 6.4 × 10^{6 }m, Density of material of earth = ρ = 5478 kg/m^{3}, G = 6.67 × 10^{-11 }N m^{2}/kg^{2}- .
**To find:**Acceleeartion due to gravity**=**g = ?

**Ans: **Acceleration due to gravity = 9.8 m/s^{2}.

#### Example – 16:

- A playful astronaut releases a bowling ball of mass 500 g into circular orbit about an altitude of 600 km. What is the mechanical energy of the ball in its orbit. radius of the earth = 6400 km, mass of earth = 6 x 10
^{24}kg. **Given:**mass of a ball = 500 g = 0.5 kg, height of satellite above the surface of the earth = 600 km, Radius of the earth = 6400 km, radius of orbit = 6400 + 600 = 7000 km = 7 x 10^{6}m, mass of earth = 6 x 10^{24}kg.**To Find:**Mechanical energy of ball = E_{T}= ?**Solution:**

Now total energy of satellite = T.E. = – B.E. = – 1.43 x 10^{7} J

**Ans:** The mechanical energy of the ball in its orbit is – 1.43 x 10^{7} J

#### Example – 17:

- Find the escape velocity from the earth for a 1000 kg spacecraft and find the kinetic energy it must have at the surface of the earth in order to escape the earth’s gravitational field.
**Given:**Mass of spacecraft = m = 1000 kg,**To Find:**Escape velocity ve = ?, kinetic energy = K.E. = ?**Solution:**

The escape velocity on the surface of the earth is given by

K.E. = ½ mv^{2} = ½ (1000)(11.2 x 10^{3})^{2}

K.E. ½ x 10^{3} x 11.2^{2} x 10^{6}

K.E.= 6.261 x 10^{10} J

**Ans:** Escape velocity is 11.2 km/s and kinetic energy required is 6.261 x 10^{10} J

Science > Physics > Gravitation > You are Here |

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