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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.
$(2.67 \times 10^{-4} N)$
Difficulty: Hard
Solution:
Mass = $m_{1} = m_{2}$= 1000 kg
Distance between the centers = d = 0.5 m
Gravitational constant = G = $6.673\times 10^{-11} Nm^{2}kg^{-2}$
Gravitational force = F = ?
F =$\frac{(G m_{1}m_{2})}{d_{2}}$
F = $6.673 \times (10) ^{-11}\times \frac{(1000 \times 1000)}{((0.5)}^{2})$
=$\frac{6.673 \times(10) ^{-11}\times (10)^{6}}{0.25}$
= $\frac{(6.673 \times (10) ^{-11}\times (10)^{6})}{0.25}$
=$\frac{(6.673 \times (10)^{-5})}{0.25}$
= $26.692\times (10) ^{-5}$ = $2.67 \times 10^{-4} N$
The gravitational force between two identical lead spheres kept at 1 m apart is 0.006673 N. Find their masses. (10, 000 kg each)
Difficulty: Hard
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.
Difficulty: Medium
Solution:
Mass of Mars = $M_{m}$ = $6.42\times 1023$ kg
Radius of Mars =$R_{m}$ = 3370 km = $3370 \times 1000$ m = $3.37 \times 10^{8} m$
Acceleration due to gravity of the surface of Mars = $g_{m}$ =?
G =$\frac{6.673*10^{-11} * 6.42*10^{23}}{11.357}$
gm =$\frac{G M_{m}}{〖R〗^{2}_{m}}$
or gm = $6.673 \times 10^{-11} \times$ $(6.42 \times 10^{23})(3.37 \times 10^{6})^{2}$
=$\frac{(6.673 \times 10^{(-11)}\times 6.42 \times 10^{23})}{11.357}$
=$\frac{42.84}{11.357}$ = 3.77 $m^{-2}$
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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.
Difficulty: Medium
Solution:
Acceleration due to gravity = $g_{m} = 1.62 ms^{-2}$
Radius of the moon = $R_{m}$ = 1740 km = $1740 \times1000$ m = $1.74 \times 10^{6} m$
Mass of moon = $M_{m}$ = ?
$g_{m}$ = $G M_{m}/〖R〗^{2}_m$
or
$M_{m}$ = $(\frac{gm\times〖R〗_{m})}{G}$
$M_{m}$ = $(\frac{1.62\times(1.74 \times10^{6}))^{2}}{(6.673 \times 10^{-11})}$
= $(\frac{1.62\times 3\times10^{12})}{(6.673\times 10^{-11})}$
$M_{m}$ = $7.35 \times 10^{22}$ kg
Calculate the value of g at a height of 3600 km above the surface of the Earth. (4.0 $ms^{-2}$)
Difficulty: Medium
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. ($0.17 \space ms^{-2}$)
Difficulty: Medium
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 $ms^{-2}$
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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)
Difficulty: Medium
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)
Difficulty: Hard
Solution:
Mass of Earth = $M_{e}$ = $6.0\times 10^{24}$ kg
Radius of Earth = $R_{e}$ = $6.4\times 10^{6}$ m
Gravitational acceleration = gh =$\frac{1}{4}$ g =$\frac{1}{4}\times 10$ $ms^{-2}$ = $2.5ms^{-2}$
Altitude above Earth’s surface = h =?
gh = $\frac{〖GM〗_{e}}{(R + h)^{2}}$
or $(R + h)^{2}$ = $\frac{〖GM〗_{e}}{g_{h}}$
Taking square root on both sides
$√16*10^{13}m^{2}– 6.4*10^{6}$
or$√((R + h)^{2})$=$\frac{√(〖GM〗_{e}}{g_{h})}$
orR + h = $\frac{√(G〖GM〗_{e}}{g_{h})}$
orh =$\frac{√(G〖 GM〗_{e}}{g_{g} -R )}$
orh = $√((\frac{6.673 \times 10^{-11}\times 6.0 \times 10^{24})}{2.5)- 6.4 \times 10^{6}}$
= $√((\frac{40.038\times 10^{13})}{2.5 )-6.4\times 10^{6}}$
= $√(16 \times 10^{13} m^{2})- 6.4\times 10^{6}$ = $√(0.16 \times 10^{12})- 6.4 \times 10^{6}$
= $0.4 \times 10^{6}- 6.4 \times 10^{6}$
= $-6.0 \times 10^{6}$ m
As height is always taken as positive, therefore
h = $6.0 \times 10^{6}$ m = One Earth’s radius
A polar satellite is launched at 850 km above Earth. Find its orbital speed.$(7431 ms^{-1})$
Difficulty: Medium
Solution:
Height = h = 850 km = $850\times 1000$ m = $0.85 \times 10^{6}$ m
Orbital velocity = $v_{o}$=?$v_{o}=\frac{√(〖 GM〗_{e}}{(R+h))}$
$v_{o}$ =$\frac{√((6.673\times 10^{-11}\times 6.0\times 10^{24})}{(6.4\times 10^{6}+0.85 \times 10^{6}))}$
=$\frac{√((40.038 \times 10^{13})}{(7.25 \times 10^{6}))}$
= $√(5.55 \times 10^{7})$ = $√(5.55 \times 10^{6}$)
= $7.431\times 10^{3}$ = $7431 ms^{-1}$
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A communication is launched at 42000 km above Earth. Find its orbital speed. $(2876 \: ms^{-1})$
Difficulty: Easy
Solution:
Height = h = 42000 km = $42000\times 1000$ m = $42 \times10^{6}$m
Orbital velocity = $v_{o}$ =?
$√8.27*10^{6}$
$v_{o}$=$\frac{√(〖 GM〗_{e}}{(R+h))}$
$v_{o}$ = $\frac{√((6.673 \times 10^{-11}\times 6.0 \times 10^{24})}{(6.4 \times 10^{6}+ 42 \times 10^{6}))}$
=$\frac{√((40.038 \times10^{13})}{48.4\times 10^{6}))}$
=$\frac{√((400.38 \times 10^{12})}{(48.4 \times 10^{6}))}$
= $√(8.27\times 10^{6})$
= $2.876 \times 10^{3} = 2876 ms^{-1}$
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