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Find the gravitational force of attraction between two spheres each of mass of 1000 kg. The distance between the centers of the spheres is 0.5 m.
$(2.67 \times 10^{-4} N)$
Solution:
The gravitational force between two identical lead spheres kept at 1 m apart is 0.006673 N. Find their masses. (10, 000 kg each)
Solution:
Gravitational force = F = 0.006673 N
Gravitational constant = G = $6.673 \times 10^{-11} Nm^{2}kg^{-2}$
Distance between the masses = d = 1m
Mass = $m_{1}$ = $m_{2}$ =?
F = G $\frac{(m_{1} m_{2})}{d_{2}}$
F = G $\frac{( m \times m )}{d_{2}}$ $(Let \: m_{1} = m_{2}= m)$
F =$\frac{(m ^{2})}{d_{2}}$
$m^{2}$ = $\frac{F \times d^{2}}{G}$
$m^{2}$ =$\frac{(0.006673 \times 〖(1) 〗^{2})}{( 6.673 \times 〖(10) 〗^{-11} )}$ =$\frac{(\frac{(6673}{1000000)}}{( 6.673 \times〖(10) 〗^{-11})}$
=$\frac{(6.673 \times〖(10) 〗^{-3})}{( 6.673 \times〖(10) 〗^{-11})}$
$√(m^{2}) = 10^{8}$
$m = 10^{4}$ = 10000 kg each
Therefore, mass of each lead sphere is 10000 kg.
Find the acceleration due to gravity on the surface of Mars. The mass of Mars is $6.42 \times 10^{23}$ kg and its radius is 3370 km.
Solution:
The acceleration due to gravity on the surface of the moon is 1.62 $ms^{-2}$. The radius of the Moon is 1740 km. Find the mass of the moon.
Solution:
Calculate the value of g at a height of 3600 km above the surface of the Earth. (4.0 $ms^{-2}$)
Solution:
Height = h = 3600 km = $3600 \times 1000$ m = $3.6 \times 10^{6}$ m
Mass of Earth = $M_{e}$ = $6.0 \times 10^{24}$ kg
Gravitational acceleration = $g_{h}$ =?
$g_{h}$ =$\frac{〖 GM〗_{e}}{(R + h)^{2}}$
$g_{h}$ = $\frac{6.673 \times 10^{-11} \times (6.0 \times 10^{24})}{((6.4 \times 10^{6} + 3.6 \times 10^{6})^{2})}$
= $\frac{6.673 \times 10^{-11}\times (6.0 \times 10^{24})}{((10.0 \times 10^{6})^{2})}$
=$\frac{6.673 \times10^{-11} \times (6.0 \times 10^{24})}{(100 \times10^{12})}$
= $6.673 \times 10^{-11}\times 6.0 \times 10^{10}$ = $40\times 10^{-1}$ = $4.0 ms^{-2}$
Find the value of g due to the Earth at geostationary satellite. The radius of the geostationary orbit is 48700 km. ($0.17 \space ms^{-2}$)
Solution:
Radius = R = $48700 \times 1000$ m = $4.87 \times 10^{7}$ m
Gravitational acceleration = g =?
g =$\cfrac{ \lparen GM _{e} \rparen}{R^{2}}$
g =$\cfrac{6.673\times 10^{-11} \times (6.0 \times10^{24})}{((4.87 \times 10^{7})^{2})}$
=$\cfrac{6.673 \times 10^{-11} \times(6.0 \times 10^{24})}{((23.717\times 10^{14})^{2})}$
=$\cfrac{(6.673\times 6.0 \times 10^{-1})}{23.717}$
=$\cfrac{4.0038}{23.717}$
= 0.17 $ms^{-2}$
The value of g is 4.0 $ms^{-2}$ at a distance of 10000 km from the center of the Earth. Find the mass of the Earth. ($5.99 \times 10^{24}$ kg)
Solution:
Gravitational acceleration = g = 4.0 $ms^{-2}$
Radius of Earth = $R_{e}$ = 10000 km = $10000 \times 1000$ m = $10^{7}$m
Mass of Earth = $M_{e}$ =?
$M_{e}$ =$\frac{(gR^{2})}{G}$
$M_{e}$ = $\frac{(4.0 \times (10^{7})^{2})}{(6.673 \times 10^{-11})}$
=$\frac{(4.0\times 10^{14})}{(6.673 \times 10^{-11})}$
= $0.599\times10^{25}$ kg = $5.99\times 10^{24}$ kg
At what altitude the value of g would become one-fourth that on the surface of the Earth? (One Earth’s radius)
Solution:
A polar satellite is launched at 850 km above Earth. Find its orbital speed.$(7431 ms^{-1})$
Solution:
A communication is launched at 42000 km above Earth. Find its orbital speed. $(2876 \: ms^{-1})$
Solution:
Height = h = 42000 km = $42000\times 1000$ m = $42 \times10^{6}$m
Orbital velocity = $v_{o}$ =?
$√8.27*10^{6}$
$v_{o}$=$\frac{√(〖 GM〗_{e}}{(R+h))}$
$v_{o}$ = $\frac{√((6.673 \times 10^{-11}\times 6.0 \times 10^{24})}{(6.4 \times 10^{6}+ 42 \times 10^{6}))}$
=$\frac{√((40.038 \times10^{13})}{48.4\times 10^{6}))}$
=$\frac{√((400.38 \times 10^{12})}{(48.4 \times 10^{6}))}$
= $√(8.27\times 10^{6})$
= $2.876 \times 10^{3} = 2876 ms^{-1}$