Gravity is nothing but a force of attraction between any two elements that exist in a universe. Elements in the universe attract each other with a specific amount of force. The mere idea of an apple falling down on the earth from a tree led to the discovery of a very interesting topic of Physics called Gravitation. Gravitation is a study of the interaction between two masses.
Question 1
Two objects attract each other with a gravitational force of magnitude \(\:1.00 \times 10^{-8} \:N\) when separated by 20.0 cm. If the total mass of the two objects is 5.00 kg, what is the mass of each?
Show Solution
Universal gravitational constant, \(\:G=6.67\times 10^{-11} \:Nm^2kg^{-2}\)
$$m_1 + m_2 = 5 \: \: \: \: (1)$$
distance of seperation between the two masses, r = 20cm = 0.2 m
The force of attraction, F between of the two masses is given by
$$F = \frac{Gm_1m_2}{r^2}$$
$$1.00\times 10^{-8} = \frac{6.67\times 10^{-11} \times m_1m_2}{0.2^2}$$
$$m_1m_2 = \frac{1.00\times 10^{-8}\times 0.04}{6.67\times 10^{-11}}$$
$$m_1m_2 = 6 \: \: \: \: \: \:(2)$$
From equation (1),
$$m_2 = 5\:- m_1 \: \: \: \: \: \: (3)$$
Putting equation (3) into equation (2), we will have
$$m_1(5-m_1) = 6$$
$$
m_{1}^2 – 5m_1 + 6=0
$$
$$(m_1-3)(m_1 – 2)=0$$
$$m_1 = 3 \: \text{or} \: m_1 = 2$$
With \(\:m_1 = 3\),
$$m_2 = 5\:- 3 = 2$$
With \(\:m_1 = 2\),
$$m_2 = 5 \: – 2 = 3$$
Therefore, the two masses are 3kg and 2 kg.
Question 2
Mt. Everest is at a height of 29,003 ft (8840 m) above sea level. The greatest depth in the sea is 34,219 ft (10,430m). Compare the Earth’s surface gravity at these two points.
Show Solution
For Mt. Everest, height \(\:h_1=8840 \: m\)
For greatest depth in the sea, height \(\:h_2=10430 \: m\)
The accegration due to gravity, \(\:g_1 \:\) on the Earth at height \(\:h_1 \:\) is given by
$$
g_1 = \frac{GM_E}{(h_1 + R_E)^2}
$$
Similarly, the accegration due to gravity, \(\:g_2 \:\) on the Earth at height \(\:h_2 \:\) is given by
$$
g_2 = \frac{GM_E}{(h_2 + R_E)^2}
$$
Thus,
$$
\frac{g_1}{g_2}=\frac{\frac{GM_E}{(h_1 + R_E)^2}}{\frac{GM_E}{(h_2 + R_E)^2}}
$$
$$
\frac{g_1}{g_2}= \left(\frac{h_2 + R_E}{h_1 + R_E} \right )^2
$$
With the radius of the earth, \(\:R_E =6.37 \times 10^6 \: m,\)
$$\frac{g_1}{g_2}= \left(\frac{10430 + 6.37 \times 10^6}{8840 + 6.37 \times 10^6} \right )^2$$
$$\frac{g_1}{g_2}=1.0005$$
Thus,the value of the acceleration due to gravity on the two locations is approximately the same as the ratio of the two values is approximatley 1.
Question 3
A satellite of mass 200 kg is launched from a site on the Equator into an orbit at 200 km above the Earth’s
surface. (a) If the orbit is circular, what is the orbital period of this satellite? (b) What is the satellite’s speed in orbit?
Show Solution
height, \(\:h=200 \: km = 200000 \: m\)
mass of satellite, \(\:m=200kg\)
Universal gravitational constant, \(\:G=6.67\times 10^{-11} \:Nm^2kg^{-2}\)
radius of the earth, \(\:R_E =6.37 \times 10^6 \: m,\)
mass of the earth, \(\:M_E =5.97 \times 10^{24} \: kg,\)
(a) The orbital period of the satellite, T is given by
$$
T=2\pi \sqrt{\frac{(h + R_E)^3}{GM_E}}
$$
$$
=6.28 \times \sqrt{\frac{(200000 + 6.37 \times 10^6)^3}{6.67\times 10^{-11} \times 5.97 \times 10^{24}}}
$$
$$=5299.8 \:\text{seconds} = \frac{5299.8}{3600} \: \text{hours}=1.5 \:\text{hours}$$
Therefore, the period of the satellite in its orbit is 1.5 hours.
(b) The satellite’s speed in orbit, \(\:v \:\) is given by
$$v=\sqrt{\frac{GM_E}{h + R_E}}$$
$$=\sqrt{\frac{6.67\times 10^{-11} \times 5.97 \times 10^{24}}{200000 + 6.37 \times 10^6}}$$
$$7785 \: m/s$$
Therefore, the satellite’s speed in orbit is 7785 m/s.
Question 4
The Sun has a mass of \(\: 1.9892 \times 10^{30} \: kg \:\) and a radius of \(\: 6.9598 \times 10^8 \: m \: \). Calculate the escape velocity from the solar surface (in km/s).
Show Solution
Universal gravitational constant, \(\:G=6.67\times 10^{-11} \:Nm^2kg^{-2}\)
mass of the sun , \(\:M_S =1.9892 \times 10^{30} \: kg,\)
radius of the earth, \(\:R_S =6.9598 \times 10^{8} \: m,\)
The escape velocity from the solar surface, \(\:v_{e} \:\) is given by
$$v_{e} = \sqrt{\frac{2GM_S}{R_S}}$$
$$v_{e} = \sqrt{\frac{2 \times 6.67\times 10^{-11} \times 1.9892 \times 10^{30}}{6.9598 \times 10^{8}}}$$
$$=617474.1 \: m/s=617.5 \: km/s$$
Question 5
The mass of the moon is \(\: 7.3 \times 10^{22} \: kg\) and its radius is \(\: 1.74 \times 10^{6} \: m\). Calculate the gravitational acceleration at its surface.
Show Solution
Universal gravitational constant, \(\:G=6.67\times 10^{-11} \:Nm^2kg^{-2}\)
radius of the moon, \(\:R_m =1.74 \times 10^{6} \: m,\)
mass of the moon, \(\:M_m =7.3 \times 10^{22} \: kg,\)
The accelration due to gravity on the moon’ surface, \(\:g_m \:\) is given by
$$g_m = \frac{GM_m}{R^2_m}$$
$$g_m = \frac{6.67\times 10^{-11} \times 7.3 \times 10^{22}}{\left ( 1.74 \times 10^{6} \right)^2}$$
$$=1.61 \: m/s^2$$
Question 6
At what height must we go so that the value of g becomes half of what it is at the surface of the earth?
Show Solution
Universal gravitational constant = G
radius of the earth, \(\:R =6.37 \times 10^{6} \: m,\)
height,h = ?
Let \(\: M \:\) be the mass of the earth and \(\:g \:\) the value of acceleration due to gravity on the earth’s surface.
Let \(\:g_h \:\) the value of acceleration due to gravity at an height of \( \: h\:\) above the earth’s surface.
Thus,
$$g = \frac{GM}{R^2}$$
$$g_h = \frac{GM}{\left( h + R\right )^2}$$
From the question,
$$g_h = \frac{1}{2}g$$
$$2\frac{GM}{\left( h + R\right )^2} = \frac{GM}{R^2}$$
$$
2R^2 = \left( h + R \right )^2
$$
$$
2R^2 = h^2 + R^2 + 2hR
$$
$$
h^2 + 2Rh – R^2 = 0
$$
Using the quadratic formular, we will have
$$
h = \frac{-2R \pm \sqrt{(2R)^2 – 4(1)(-R^2)}}{2(1)}
$$
$$
= \frac{-2R \pm \sqrt{4R^2 + 4R^2}}{2}
$$
$$
h = \frac{-2R \pm \sqrt{8R^2}}{2}
$$
$$
h = \frac{-2R \pm R\sqrt{8}}{2}
$$
Taking the positive value, we will have
$$
h = \frac{-2R + R\sqrt{8}}{2}
$$
$$
h = R\left(\frac{-2 + \sqrt{8}}{2} \right )
$$
$$
h = 6.37 \times 10^{6} \times \left(\frac{-2 + \sqrt{8}}{2} \right )
$$
$$=2.64 \times 10^6 \: m$$
Therefore, the height at which the acceleration due to gravity becomes half is \(\: 2.64 \times 10^6 \: m\)
