Question

Solve the following examples.

a. An object takes 5 s to reach the ground from a height of 5 m on a planet. What is the value of g on the planet?

b.The radius of planet A is half the radius of planet B. If the mass of A is M_A, what must be the mass of B so that the value of g on B is half that of its value on A?

c. The mass and weight of an object on earth are 5 kg and 49 N respectively. What will be their values on the moon? Assume that the acceleration due to gravity on the moon is 1/6th of that on the earth.

d. An object thrown vertically upwards reaches a height of 500 m. What was its initial velocity? How long will the object take to come back to the earth? Assume g = 10 m/s²

e. A ball falls off a table and reaches the ground in 1 s. Assuming g = 10 m/s², calculate its speed on reaching the ground and the height of the table.

f. The masses of the earth and moon are 6 × 10²⁴ kg and 7.4 × 10²² kg, respectively. The distance between them is 3.8 × 10⁵ km. Calculate the gravitational force of attraction between the two? Use G = 6.7 × 10^–11 N m² kg^–2

g. The mass of the earth is 6 × 10²⁴ kg. The distance between the earth and the sun is 1.5 × 10¹¹ m. If the gravitational force between the two is 3.5 × 10²² N, what us the mass of the sun? Use G = 6.7 × 10^–11 N m² kg^–2

Answer 1

a. Here, u = 0
S = 5 m
t = 5 s
From second equation of motion, we have
$$\begin{gathered} S=ut+\frac{1}{2}g{t}^2 \\ 5=\frac{1}{2}g(5{)}^2 \\ g=\frac{10}{25}=0.4\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$
Hence, the value of g on the planet is 0.4 m/s².

b. The acceleration due to gravity of a planet is given as
$g=\frac{GM}{r^2}$
For planet A: $g_{\mathrm{A}}=\frac{G{M}_{\mathrm{A}}}{r_{\mathrm{A}}^2}$
For planet B: $g_{\mathrm{B}}=\frac{G{M}_{\mathrm{B}}}{r_{\mathrm{B}}^2}$
Now,
$$\begin{gathered} g_{\mathrm{B}}=\frac{1}{2}{g}_{\mathrm{A}}\left(\mathrm{Given}\right) \\ \mathrm{or},\frac{G{M}_{\mathrm{B}}}{r_{\mathrm{B}}^2}=\frac{G{M}_{\mathrm{A}}}{2r_{\mathrm{A}}^2} \\ \Rightarrow M_{\mathrm{B}}=\frac{M_{\mathrm{A}}{r}_{\mathrm{B}}^2}{2r_{\mathrm{A}}^2} \end{gathered}$$
Given: $r_{\mathrm{A}}=\frac{1}{2}{r}_{\mathrm{B}}$
$\Rightarrow M_{\mathrm{B}}=\frac{M_{\mathrm{A}}{r}_{\mathrm{B}}^2}{2({\displaystyle \frac{1}{2}}{r}_{\mathrm{B}}{)}^2}=2M_{\mathrm{A}}$
Thus, the mass of planet B should be twice that of planet A.

c. Mass of the object on Earth, m = 5 kg
Weight of the object on Earth, W_E = 49 N
Weight of the object on Moon,
$$\begin{gathered} W_{\mathrm{M}}=\frac{1}{6}{W}_{\mathrm{E}} \\ \Rightarrow W_{\mathrm{M}}=\frac{49}{6}=8.17\mathrm{N} \end{gathered}$$
Mass of the object on Moon = 5 kg (since mass is independent of the place of observation)

d. For vertical upward motion of the object,
S = 500 m
g = $-$10 m/s²
v = 0
Let u be the initial velocity of the object. From third equation of motion, we have
$$\begin{gathered} v^2-u^2=2aS \\ \Rightarrow 0-u^2=-2\times 10\times 500 \\ \Rightarrow u=100\mathrm{m}/\mathrm{s} \end{gathered}$$
Now, let t₁ be time taken by the object to reach at 500 m height. Thus,
$$\begin{gathered} v=u+at \\ \Rightarrow 0=100-10\times t_1 \\ {t}_1=10\mathrm{s} \end{gathered}$$
For vertical downward motion of the object,
S = 500 m
g = 10 m/s²
u = 0
Let t₂ be the time taken by the object to come back to the Earth from height of 500 m.
From second equation of motion, we have
$$\begin{gathered} S=ut+\frac{1}{2}a{t}^2 \\ \Rightarrow 500=\frac{10}{2}\times t_2^2 \\ \Rightarrow t_2=10\mathrm{s} \end{gathered}$$
Thus, the total time taken by the object to reach back to Earth = t₁ + t₂ = 20 s

e. Here, t =1 s
g = 10 m/s²
u = 0
Let v be the velocity of the ball on reaching the ground.
Thus, from first equation of motion, we have
v = u + gt
$\Rightarrow$v = 10$\times$1 = 10 m/s
Hence, the speed of the object on reaching the ground is 10 m/s.
Let h be the height of the table. Thus, from second equation of motion, we have
$$\begin{gathered} S=ut+\frac{1}{2}g{t}^2 \\ \Rightarrow h=0+\frac{1}{2}\times 10\times 1^2 \\ \Rightarrow h=5\mathrm{m} \end{gathered}$$
Hence, the height of the table is 5 m.

f. The gravitational force between the Moon and the Earth can be found out using the formula,
$F=G\frac{M_{\mathrm{e}}{M}_{\mathrm{m}}}{R^2}$
where, M_e and M_m are the masses of the Earth and the Moon, respectively. Using all the given values, we have
$F=(6.7\times {10}^{-11})\frac{(6\times {10}^{24})(7.4\times {10}^{22})}{(3.8\times {10}^5\times 1000{)}^2}=20.6\times {10}^{19}\simeq 2\times {10}^{20}\mathrm{N}$

g. The gravitational force between the Sun and the Earth can be found out using the formula,
$F=G\frac{M_{\mathrm{e}}{M}_{\mathrm{s}}}{R^2}$
where, M_e and M_s are the masses of the Earth and the Sun, respectively. Using all the given values, we have
$$\begin{gathered} 3.5\times {10}^{22}=(6.7\times {10}^{-11})\frac{(6\times {10}^{24})\times M_{\mathrm{s}}}{(1.5\times {10}^{11}{)}^2} \\ \Rightarrow M_{\mathrm{s}}=\frac{(3.5\times {10}^{22})\times (1.5\times {10}^{11}{)}^2}{(6.7\times {10}^{-11})\times (6\times {10}^{24})}=1.96\times {10}^{30}\mathrm{kg} \end{gathered}$$