3 and 4 Marks Questions
 Mass of the Jupiter is 1.9 × 10^{30} kg and mass of the Sun is 2 × 10^{30} kg. Calculate the distance between them if the gravitational force between them is 4.3 × 10^{29} N.

Here,
Mass of the Jupiter (m_{1}) = 1.9 × 10^{30} Kg
Mass of the Sun (m_{2}) = 2 × 10^{30} kg
Gravitational force (F) = 4.3 × 10^{29} N
Distance between Jupiter and the Sun (d) = ?We know,
∴ Distance between Jupiter and the Sun (d) = 2.94 × 10^{10} m.
F = $\frac{Gm_{1}m_{2}}{d^2}$
or, 4.3 × 10^{29} = $\frac{6.67 \times 10^{11} \times 1.9 \times 10^{30} \times 2 \times 10^{30}}{d^2}$
or, d^{2} = $\frac{25.346 \times 10^{49}}{4.3 \times 10^{29}}$
or, d^{2} = 5.894 × 10^{20}  The gravitational force produced between any two objects kept 2.5×10^{4} km apart is 580N. At what distance should they be kept so that the gravitational force becomes half.

Here,
Gravitational force (W_{1}) = 580N
W_{2} = 580/2 = 290N
distance (R_{1}) = 2.5 × 10^{4} km = 2.5 × 10^{7}m
R_{2} = ?Now, we know,
$\frac{W_1}{W_2}$ = $\frac{(R_2)^2}{(R_1)^2}$
or, $(R_2)^2$ = $\frac{W_1 \times (R_1)^2}{W_2}$
= $\frac{580 \times (2.5 \times 10^7)^2}{290}$
= ${2 \times 6.25 \times 10^{14}}$
= ${12.5 \times 10^{14}}$∴ R_{2} = $\sqrt{12.5 \times 10^{14}}$ = 3.54 × 10^{7}m.
 A stone dropped freely from 72 m height of the tower and reaches the ground in 6 seconds. Calculate acceleration due to gravity of that stone. At what condition does the acceleration due to gravity becomes zero.

Here,
Height (h) = 72m
Initial velocity (u) = 0 m/s
time taken (t) = 6 s
acceleration due to gravity (g) = ?
Now, we have,
from equation of motion,
h = ut + $\frac{1}{2}$gt^{2}
or, 72 = 0 × 6 + $\frac{1}{2}$ × g × 6 × 6
or, g = $\frac{72 \times 2}{36}$
∴ g = 4 m/s^{2}
When the object is brought to the center of the earth, the value of acceleration due to gravity is zero.  How much time does a stone take to reach on the surface of the earth drop from the height 20m and what will be its acceleration due to gravity after 3 seconds? (g = 10 m/s^{2})

Here,
height (h) = 20m
acceleration due to gravity (g) = 10 m/s^{2}
time taken (t) = ?
Now, we have,
From equation of motion,
h = ut + $\frac{1}{2}$gt^{2}
or, 2h = gt^{2} (u = 0)
or, t^{2} = $\frac{2h}{g}$
= $\frac{2 \times 20}{10}$
∴ t = 2 second
Its acceleration due to gravity after 3 seconds is 0 m/s^{2}. 
In the figure below, two identical metal balls A and B having equal masses
are being dropped towards the surface of moon and earth. Analyze the given
data and answer the following questions:

 If both the metal balls are released simultaneously, which one does strike the ground faster? Show with calculation.

For the ball falling on the earth (B):
From equation of motion,
h = ut + $\frac{1}{2}$gt^{2}
or, 2h = gt^{2} (u = 0)
or, t = $\sqrt{\frac{2h}{g}}$
= 1.43 secFor the ball falling on the moon (A)
∴ The ball B takes short time to strike the ground i.e. B strikes faster than A.
t = $\sqrt{\frac{2 \times 10}{1.62}}$
= 3.51 sec  A person lifts 30kg on the surface of the earth. How much mass can he lift on the surface of the moon if he applies same magnitude of force?

According to enquestion,
weight lifted on the earth = weight lifted on the moon
i.e. m_{1} × g_{e} = m_{2} × g_{m}
or, 30 × 9.8 = m_{2} × 1.62
or, m_{2} = $\frac{30 \times 9.8}{1.62}$
∴ m_{2} = 181.48 kg
Thus, the person can lift 181.48 kg on the surface of the moon.
 The mass of the moon is 7.2×10^{22} kg and its radius 1.7×10^{6} m. Calculate the acceleration due to gravity of the moon and also calculate the weight of an object of mass 80 kg on the lunar surface.

Here,
Mass of the moon (M) = 7.2 × 10^{22} kg
Radius (R) = 1.7 × 10^{6} m
Gravitational constant (G) = 6.67 × 10^{11} Nm^{2}/kg^{2}
Acceleration due to gravity (g) = ?Now, we know,
Finally, mass of an object (m) = 80 kg
g = $\frac{GM}{R^2}$
= $\frac{6.67 \times 10^{11} \times 7.2 \times 10^{22}}{(1.7 \times 10^6)^2}$
∴ g = 1.67 m/s^{2}
weight of an object (W) = m × g = 80 × 1.67 = 133.6 N  A meteor is falling towards the Earth. If mass and radius of the earth are 6×10^{24} kg and 6.4×10^{3} km respectively. Find the height of meteor from the earth's surface where its acceleration due to gravity becomes 4 m/s^{2}. The weight of a body decreases in a coal mine, why?

Here,
mass of Earth (M) = 6 × 10^{24} kg
Radius (R) = 6.4 × 10^{3} km = 6.4 × 10^{6} m
acceleration due to gravity (g) = 4m/sec^{2}
Gravitational constant (G) = 6.67 × 10^{11}Nm^{2} /kg^{2}Now, g = $\frac{GM}{(R+h)^2}$
Since coal mining takes place at depths, the value of g decreases at a height less than the surface of the earth. Therefore, the weight of the object in the coal mine is reduced.
or, g = $\frac{6.67 \times 10^{11} \times 6 \times 10^{24}}{(6.4 \times 10^6 + h)^2}$
or, (6.4 × 10^{6} + h)^{2} = 10.005 × 10 ^{18}
∴ h = 3.6 × 10^{6}  The gravitational force produced between two bodies is 25N when they are at the distance of 4m. How much gravitational force is produced when they are kept 2m apart? Calculate.

Here,
1st case,
Gravitational Force (F_{1}) = 25N
Distance between two bodies(d_{1}) = 4m
Now, we know,
F_{1} = $\frac{GM_{1}M_{2}}{d^2}$
Or, 25 × 16 = GM_{1}M_{2}  (I)Again, 2nd case,
Hence, 100N gravitational force is produced when they are kept 2m apart.
Gravitational Force (F_{2}) = ?
Distance between two bodies(d_{2}) = 2m
Now,
F_{2} = $\frac{GM_{1}M_{2}}{d^2}$
= $\frac{25 \times 16}{4}$ (from equation I)
∴ F_{2} = 100N  The mass and radius of the moon are 7.1 × 10^{22} kg and 1.7 × 10^{3} km respectively. Calculate the acceleration due to gravity on the moon's surface.

Here,
Mass of moon (M) = 7.1 × 10^{22} kg
Radius of moon (R) = 1.7 × 10^{3} km = 1.7 × 10^{6} m
Acceleration due to gravity (g) = ?Now, we know,
g = $\frac{GM}{R^2}$
= $\frac{6.67 \times 10^{11} \times 7.1 \times 10^{22}}{(1.7 \times 10^6 )^2}$
Hence. The acceleration due to gravity of the moon’s surface is 1.63 m/s^{2}.
=$\frac{16.386}{10}$
= 1.63 m/s^{2}