Numerical Problems

Oct 12, 2022

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 NGravitational constant = G = $6.673 \times 10^{-11} Nm^{2}kg^{-2}$Distance between the masses = d = 1mMass = $m_{1}$ = $m_{2}$ =?        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 eachTherefore, 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 $m^{-2}$

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 = $4.87 \times 10^{7}$ mGravitational 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)

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 = $v_{o}$=?$v_{o}=\frac{√(〖 GM〗_{e}}{(R+h))}$   $v_{o}$ =$\frac{√((6.673\times 10^{-11}\times 6.0\times 10^{24})}{(6.4\times 10^{6}+0.85 \times 10^{6}))}$ =$\frac{√((40.038 \times 10^{13})}{(7.25 \times 10^{6}))}$       = $√(5.55 \times 10^{7})$ = $√(5.55 \times 10^{6}$)       = $7.431\times 10^{3}$ = $7431 ms^{-1}$

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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