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Science > Physics > Gravitation > You are Here |

#### Example – 01:

- Show that the critical velocity of a body revolving in a circular orbit very close to the surface of the planet of radius R and mean density ρ is

**Solution:**

The critical velocity of a satellite orbiting very close to the earth’s surface is given by

- This is the expression for the critical velocity of a satellite orbiting very close to the earth’s surface in terms of density of the material of the planet.

#### Example – 02:

- What would be the duration of the year if the distance between the earth and sun gets doubled the present distance?
**Solution:****Given:**r_{2}= 2 r_{1}, old period T_{1}= 365 days**To Find:**New period T_{2}=?

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

**Ans:** Thus the duration of the year would be 1032 days

#### Example – 03:

- Calculate the height of communication satellite above the surface of the earth? Given G = 6.67 x 10
^{-11}N m^{2}/kg^{2}; radius of earth = 6400 km, Mass of the earth = 6 x 10^{24}Kg - Given: For communication satellite, T = 24 hr = 24 x 60 x 60 s, Universal gravitational Constant = G = 6.67 x 10
^{-11}N m^{2}/kg^{2}; radius of earth = R = 6400 km = 6.4 x 10^{6}m, Mass of the earth = M = 6 x 10^{24}Kg - To find: height of communication satellite above the surface of the earth = h = ?

The time period of satellite is given by

Now, r = R + h

Hence, h = r – R = – 6.37 x 106

= 35.95 x 106 m

= 35.95 x 103 x 103 m

= 35950 km

**Ans:** The height of satellite above the surface of the earth is 35950 km.

#### Example – 04:

- The distances of two planets from the sun are 10
^{13}m 10^{12}m respectively. Find the ratio of their periods and orbital speeds. **Given:**r_{1}= 10^{13}m , r_{2}= 10^{12}m**To find:**T_{1}: T_{2}= ?, v_{c1}: v_{c2}= ? ,

**Ans : **The ratio of their period is 31.62:1

The ratio of their critical velocities is 0.3162:1

#### Example – 0.5:

- A body weights 3.5 kg wt, on the surface of the earth. How much will it weigh on the surface of a planet whose mass is 1/7 that of earth and whose radius is half that of earth
**Solution****:****Given:**, Weight of body on earth = W_{E}= 1.8 kg, mass of planet = 1/9 mass of earth i.e M_{P}= 1/7 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 – 06:

- The radii of the orbits of two satellites revolving around the earth are in the ratio 3:8. Compare their i) critical speed and ii) periods.
**Given:**ratio of radii of orbits r_{1}: r_{2 }= 3:8**To find:**v_{c1}: v_{c2}=? , T_{1}: T_{2}= ?

**Ans : **The ratio of their critical velocities is 1.633:1

The ratio of their period is 0.2296:1

#### Example – 07:

- Calculate the escape velocity of a body from the surface of the earth, Given: G = 6.67 x 10
^{-11}N m^{2}/kg^{2}, Radius of the earth = 6400 km, and g =9.8 m/s^{2}. - Solution:
**Given:**radius of earth = R = 6400 km = 6.4 x 10^{6}m, Acceleration due to gravity = 9.8 m/s^{2}. Universal gravitational Constant = G = 6.67 x 10^{-11}N m^{2}/kg^{2},**To Find:**v_{e}=?

**Ans: **Thus the escape velocity of a body from the surface of the earth is 11.2 km/s.

#### Example – 08:

- Find the binding energy of a body of mass 50 kg at rest on the surface of the earth. Given: G = 6.67 x 10
^{-11}N m^{2}/kg^{2}, R = 6400 km = 6.4 x 10^{6}m; M = 6 x 10^{24}kg. **Given:**the mass of body = 50 kg, Universal gravitational constant = G = 6.67 x 10^{-11}N m^{2}/kg^{2}, R = 6400 km = 6.4 x 10^{6}m; M = 6 x 10^{24}kg.**To find:**binding energy of satellite = B.E. =?**Solution:**

The binding energy of the body at rest on the surface of earth is given by

**Ans:** Binding energy of the body is 3.127 x 10^{9} J

#### Example – 09:

- What are the total energy and binding energy of an artificial satellite of mass 1000 kg orbiting at a height of 1600 km above the surface of the earth? Mass of the earth = 6 x 10
^{24}kg; Radius of earth = 6400 km, G = 6.67 x 10^{-11}N m^{2}/kg^{2}, - Given: m = 100 kg, Universal gravitational Constant = G = 6.67 x 10
^{-11}N m^{2}/kg^{2}, R = 6400 km = 6.4 x 10^{6}m; M = 6 x 10^{24}kg. h = 1600 km, r = R + h = 6400 + 1600 = 8000 km = 8 x 10^{6}m; - To Find: T.E. = ?, B.E. = ?,
**Solution:**

Now Total energy = – B.E. = – 2.5 x 10^{10} J

**Ans: **Total energy of satellite = – 2.5 x 10^{10} J and

The binding energy of satellite = 2.5 x 10^{10} J

#### Example – 10:

- What would be the duration of the year if the distance between the earth and sun gets halved the present distance?
**Solution:****Given:**r_{2}= 1/2 r_{1}, old period T_{1}= 365 days**To Find:**New period T_{2}=?

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

**Ans:** Thus duration of the year would be 129 days

#### Example – 11:

- Two bodies of masses 5 kg and 6 x 10
^{24}kg are placed with their centres 6.4 x 10^{6}m apart. Calculate the force of attraction between the two masses. Also, find the initial acceleration of two masses assuming no other forces act on them. **Solution:****Given:**Mass of first body = m_{1}= 5 kg, mass of second body = m_{2}= 6 x 10^{24}kg, Distance between masses = r = 6.4 x 10^{6}m, Universal gravitational Constant = G = 6.67 x 10^{-11}N m^{2}/kg^{2}.**To Find:**Force of attraction between two masses = F = ?, Initial accelerations of the two masses =?

By Newton’s law of gravitation

- Initial acceleration of body of mass 5 kg

By Newton’s second law of motion F = ma

Thus a = F/m = 48.85 / 5 = 9.77 m/s^{2}

- Initial acceleration of body of mass 6 x 10
^{24}kg

By Newton’s second law of motion F = ma

Thus a = F/m = 48.85 / 6 x 10^{24 }= 8.142 x 10^{-24 }m/s^{2}

Ans: The force of attarction between the two masses = 48.85 N

The Initial acceleartion of body of mass 5 kg is 9.77 m/s^{2 }and

That of body of mass 6 x 10^{24} kg is 8.14 x 10^{-24 }m/s^{2}.

**Example – 12:**

- Find the value of Universal gravitational Constant G from the following data: M = 6 x 10
^{24}kg, R = 6400 km, g = 9.774 m/s^{2},

**Given:**M = 6 x 10^{24 }kg, R = 6400 km = 6.4 x 10^{6}m, g = 9.774 m/s^{2},

**To find:**G = ? **Solution:**

**Ans:**The value of G is 6.672 x 10^{-11}N m^{2}/kg^{2}.

#### Example – 13:

- Assuming the earth to be homogeneous sphere, find the density of material of the earth from the following data.g = 9.8 m/s
^{2}, G = 6.673 x 10^{-11}Nm^{2}/kg^{2}, R = 6400 km, **Given:**g = 9.8 m/s^{2}, Universal gravitational Constant = G = 6.673 x 10^{-11}Nm^{2}/kg^{2},R = 6400 km = 6.4 x 10^{6}m,**To find:**Density = r = ?**Solution:**

**Ans:** Density of material of the earth = 5483 kg/m^{3}

#### Example – 14:

- The mass of the body on the surface of the earth is 100 kg. What will be its mass and weight at an altitude of 1000 km.
**Solution****:****Given:**, Mass of body = 100 kg, R = 6400 km and altitude = h = 1000 km**To find:**Mass and weight at altitude of 1000 km = ?

r = R + h = 6400 + 1000 = 7400 km

Now weight of body at altitude 100 km = W_{h} = m g_{h} = 100 x 7.33 = 733 N

The mass is always constant, hence mass at an altitude of 100 km = 100 kg

**Ans: **At an altitude of 1000 km the mass of the body is 100 kg and its weight is 733 N.

Science > Physics > Gravitation > You are Here |

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