9th Class Physics Unit No. 5 Gravitation (Numerical Problems with Solutions)

 

UNIT No. 5

Gravitation

(Solution of Numerical Problems)

5.1         Find the gravitational force of attraction between two spheres each of mass 1000 kg. The distance between the centres of the spheres is 0.5 m.

Data

Mass of 1^st sphere = m₁ = 1000 kg

Mass of 1^st sphere = m₁ = 1000 kg

Distance between the centre of the spheres = d = 0.5 m

Gravitational constant = G = 6.67x10^-11 Nm²kg^-2

Gravitational force of attraction between the spheres = F = ?

Formula

$$F = G\frac{m_{1}m_{2}}{d^{2}}$$

Solution

 

$$F = G\frac{m_{1}m_{2}}{d^{2}}$$

         Putting values in this equation

$$F = 6.67x10^{-11}\frac{(1000)(1000)}{(0.5)^{2}}$$

F= 2.67x10^-4 N

 

5.2        The gravitational force between two identical lead spheres kept at 1 m apart is 0.006673 N. Find their masses.

Data

Gravitational force between the spheres = F = 0.006673 N

Distance between the centre of the spheres = d = 0.5 m

Gravitational constant = G = 6.67x10^-11 Nm²kg^-2

Mass of 1^st sphere = m₁ = ?

Mass of 1^st sphere = m₁ = ?

Formula

$$F = G\frac{m_{1}m_{2}}{d^{2}}$$

Solution

$$F = G\frac{m_{1}m_{2}}{d^{2}}$$

         Solve the formula for the value of m₁ m₂

$$m_{1}m_{2} = \frac{Fd^{2}}{G}$$

As both the masses are identical therefore

m₁ = m_{2 =} m

Then the above formula becomes

$$m^{2} = \frac{Fd^{2}}{G}$$

Taking square root on both sides

$$m = \sqrt{\frac{Fd^{2}}{G}}$$

Putting values in this equation

$$m = \sqrt{\frac{(0.006673))(0.5))^{2}}{6.67x10^{-11}}}$$

m = 1000 kg

Because both the spheres are identical.

Mass of 1^st sphere = m₁ = 1000 kg

Mass of 1^st sphere = m₁ = 1000 kg

5.3        Find the acceleration due to gravity on the surface of the Mars. The mass of Mars is 6.42x10²³ kg and its radius is 3370 km.

Data

Mass of the Mars = M_M = 6.42x10²³ kg

Radius of the Mars = R_M = 3370 km = 3370x10³ m

Gravitational constant = G = 6.67x10^-11 Nm²kg^-2

Acceleration due to gravity on the surface of the Mars = g_M = ?

Formula

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

Solution

$$g_{M} = \frac{GM_{M}}{R_{M}^{2}}$$

Putting values in this equation

$$g_{M} = \frac{(6.67X10^{-11})(6.42X10^{23}))}{3370X10^{3}}$$

g = 3.77 ms^-1      Ans

5.4        The acceleration due to gravity on the surface of moon is 1.62 ms^-2. The radius of moon is 1740 km. Find the mass of moon.

Data

Acceleration due to gravity on the surface of the Moon = g_m = 1.62 ms^-2

Radius of the Moon = R_m = 1740 km = 1740x10³ m

Gravitational constant = G = 6.67x10^-11 Nm²kg^-2

Mass of the Mars = M_m = ?

Formula

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

Solution

$$g_{m} = \frac{GM_{m}}{R_{m}^{2}}$$

rearranging this equation for the value of M_m

$$M_{m} = \frac{GR_{m}^{2}}{g_{m}}$$

_{Putting values in this equation}

$$M_{m} = \frac{(1.62)(1.740X^{6})^{2}}{6.67X10^{-11}}$$  

M_m = 7.35x10²³ kg        Ans

5.5        Calculate the value of g at a height of 3600 km above the surface of the Earth.

Data

Height above the surface of earth = h = 3600 km = 3.6x10⁶ m

Radius of the earth = R = 6400 km = 6.4x10⁶ m

Mass of the earth = M = 6x10²⁴ kg

Gravitational constant = G = 6.67x10^-11 Nm²kg^-2

Acceleration due to gravity above the surface of earth = g_h = ?

Formula

$$g_{h} = \frac{GM}{(R+h)^{2}}$$

Solution

$$g_{h} = \frac{GM}{(R+h)^{2}}$$

         Putting values in this equation

$$g_{h} = \frac{(6.67X10^{-11})(6X10^{24})}{(6.4X10^{6}+3.6X10^{6})^{2}}$$  

g_h = 4 ms^-2         Ans

5.6        Find the value of g due to the Earth at geostationary satellite. The radius of the geostationary orbit is 48700 km.

Data

Radius of the geo stationary orbit =  R = h = 48700 km = 4.8x10⁷ m

Mass of the earth = M = 6x10²⁴ kg

Gravitational constant = G = 6.67x10^-11 Nm²kg^-2

Acceleration due to gravity at the geostationary satellite = g_h = ?

Formula

$$g_{h} = \frac{GM}{(R+h)^{2}}$$

Solution

$$g_{h} = \frac{GM}{(R+h)^{2}}$$

          Putting values in this equation

$$g_{h} = \frac{(6.67X10^{-11})(6X10^{24})}{(4.8X^{7})^{2}}$$  

g_h = 0.17 ms^-2    Ans

 

5.7        The value of g is 4.0 ms^-2 at a distance of 10000 km from the centre of the Earth. Find the mass of the Earth.

Data

Acceleration due to gravity above the surface of earth = g_h = 4 ms^-2

Distance from the centre of  earth = R+h = 10000 km = 1x10⁷ m

Gravitational constant = G = 6.67x10^-11 Nm²kg^-2

Mass of the Mars = M = ?

Formula

$$g_{h} = \frac{GM}{(R+h)^{2}}$$

Solution

$$g_{h} = \frac{GM}{(R+h)^{2}}$$

rearranging this equation for the value of M

$$M = \frac{g_{h}(R+h)^{2}}{G}$$

putting values in this equation

$$M = \frac{(4)(1X^10{7})^{2}}{6.67X10^{-11}}$$

M = 5.99x10²⁴ kg        Ans

 

5.8      At what altitude the value of g would become one fourth than on the surface of the Earth?

Data

Acceleration due to gravity above the surface of earth = g_h = g/h = 9.8/4 = 2.45  ms^-2

Radius of earth = h = 6400 km = 6.4x10⁶ m

Gravitational constant = G = 6.67x10^-11 Nm²kg^-2

Mass of the earth = M = 6x10²⁴ kg

Height above the surface of  earth = h =  ?

Formula

$$g_{h} = \frac{GM}{(R+h)^{2}}$$

Solution

$$g_{h} = \frac{GM}{(R+h)^{2}}$$

rearranging this equation for the value of (R+h)²

$$(R+h)^{2}= \frac{GM}{g_{h}}$$

        Taking square root on both sides of the above equations

$$(R+h)= \sqrt{\frac{GM}{g_{h}}}$$

        rearranging this equation for the value of h

$$h= \sqrt{\frac{GM}{g_{h}}}+R$$

putting values in this equation

$$h= \sqrt{\frac{(6.67X10^{-11})(6X10^{24})}{2.45}}+6.4X10^{6}$$

h = 6.38x10⁶ m Ans (This height is one earth’s radius approximately)

5.9        A polar satellite is launched at 850 km above Earth. Find its orbital
speed.

Data

Height of satellite above the surface of earth = h = 850 km = 8.5x10⁵ m

Radius of the earth = R = 6400 km = 6.4x10⁶ m

Mass of the earth = M = 6x10²⁴ kg

Gravitational constant = G = 6.67x10^-11 Nm²kg^-2

Orbital speed of satellite = v_o = ?

Formula

$$v_{o}= \sqrt{g_{h}(R+h)}$$

$$g_{h} = \frac{GM}{(R+h)^{2}}$$

Solution

$$v_{o}= \sqrt{g_{h}(R+h)}$$      ----------- (1)

 

$$g_{h} = \frac{GM}{(R+h)^{2}}$$  -------------- (2)

 Putting value of g_h from equation (2) in equation (1)

$$v_{o}= \sqrt{\frac{GM}{(R+h)^{2}}(R+h)}$$

$$v_{o}= \sqrt{\frac{GM}{(R+h)}}$$

Putting values in this equation

$$v_{o}= \sqrt{\frac{{(6.67X10^{-11})(6X10^{24})}}{6.4X10^{6}+8.5X10^{6}}}$$

v_o = 7430 ms^-1   Ans

 

5.10   A communication satellite is launched at 42000 km above Earth. Find its orbital speed.

Data

Height of satellite above the surface of earth = h = 42000 km = 4.2x10⁷ m

Radius of the earth = R = 6400 km = 6.4x10⁶ m

Mass of the earth = M = 6x10²⁴ kg

Gravitational constant = G = 6.67x10^-11 Nm²kg^-2

Orbital speed of satellite = v_o = ?                                

 

Formula

$$v_{o}= \sqrt{g_{h}(R+h)}$$

$$g_{h} = \frac{GM}{(R+h)^{2}}$$

Solution

$$v_{o}= \sqrt{g_{h}(R+h)}$$        ----------- (1)

$$g_{h} = \frac{GM}{(R+h)^{2}}$$  -------------- (2)

          Putting value of g_h from equation (2) in equation (1)

$$v_{o}= \sqrt{\frac{GM}{(R+h)^{2}}(R+h)}$$

$$v_{o}= \sqrt{\frac{GM}{(R+h)}}$$

Putting values in this equation

$$v_{o}= \sqrt{\frac{{(6.67X10^{-11})(6X10^{24})}}{6.4X10^{6}+4.2X10^{6}}}$$

v_o = 2876 ms^-1   Ans