Question

Good and challenging question here ?


•Spheres A and B of uniform density have masses 1 kg and 100 kg respectively.
• Their centres are separated by 100 m.
(i) Find the gravitational force between them.
(ii) Find the gravitational force on A due to the earth.
(iii) Suppose A and B are initially at rest and A can move freely towards B.
What will be the velocity of A one second after it starts moving towards B?
How will this velocity change with time? How much time will A take to move towards B by 1 cm?
(iv) If A begins to fall, starting from rest, due to the earth's downward pull, what will be its velocity after one second?
How much time will it take to fall through 1 cm?

[M(earth) = 6 x 1024 kg, earth) = 6400 km]​​

Answer 1

Answer:

Mass of the sphere A, m₁ = 1 kg

Mass of the sphere B, m₂ = 100 kg

Distance between them, d = 100 m

Mass of the earth, M = 6 × 10²⁴ kg

Radius of the earth, R = 6400 km = 6400 × 10³ m

Gravitational constant, G = 6.67 × 10⁻¹¹ N-m²/kg²

(i) The gravitational force between A and B :

$\sf F=\dfrac{Gm_1 m_2}{d^2}$

$\longrightarrow \rm F=\dfrac{6.67 \times 10^{-11} \times 1 \times 100}{100^2} \\\\ \longrightarrow \rm F=6.67 \times 10^{-13} \: N$

(ii) The gravitational force on A due to the earth :

$\sf F_A=\dfrac{GMm_1}{R^2}$

$\longrightarrow \rm F_A =\dfrac{6.67 \times 10^{-11} \times 1 \times 6 \times 10^{24}}{(6400 \times 10^3)^2} \\\\ \longrightarrow \rm F_A =\dfrac{6.67 \times 10^{-11} \times 6 \times 10^{24}}{64 \times 64 \times 10^{10}} \\\\ \longrightarrow \rm F_A =\dfrac{6.67 \times 6}{4096} \times 10^{3} \\\\ \longrightarrow \rm F_A =0.00977 \times 10^3 \\\\ \longrightarrow \rm F_A =9.77 \: N$

(iii) initial velocity, u = 0 m/s

Using v = u + at

Here, acceleration of the sphere A, a = F/m₁

Substituting the values,

$\longrightarrow \rm v=0+\dfrac{6.67 \times 10^{-13}}{1} (1) \\\\ \longrightarrow \rm v=6.67 \times 10^{-13} \: m/s$

∴ The velocity of A one second after it starts moving towards B is 6.67 × 10⁻¹³ m/s

⇒ A is moving towards B. The separation between A and B decreases with time and hence, the force between them increases. Then, the acceleration increases. As v = at, if acceleration increases the velocity also increases.

∴ The velocity of A will increase with time.

⇒ To find the time taken by A to move towards B by 1 cm ;

$\tt S=ut+\dfrac{1}{2}at^2$

Substituting the values,

$\longrightarrow \rm 0.01=0(t)+\dfrac{1}{2}\bigg(\dfrac{6.67 \times 10^{-11}}{1}\bigg) t^2 \\\\ \longrightarrow \rm 0.01=3.335 \times 10^{-11} \times t^2 \\\\ \longrightarrow \rm t^2=\dfrac{0.01}{3.335 \times 10^{-11}} \\\\ \longrightarrow \rm t^2 \approx 0.3 \times 10^{11} \\\\ \longrightarrow \rm t=1.732 \times 10^5 \: s$

∴ The time taken by A to move towards B by 1 cm is 1.732 × 10⁵ s

(iv) initial velocity, u = 0 m/s

Using v = u + at

Here, acceleration of the sphere A,

$\sf a=\dfrac{F_A}{m_1}$  as it starts to fall due to the earth's downward pull.

Substituting the values,

$\longrightarrow \rm v=0+\dfrac{9.77}{1} (1) \\\\ \longrightarrow \rm v=9.77\: m/s$

∴ The velocity of A after one second will be 9.77 m/s.

⇒ To find the time taken by A to fall through 1 cm ;

using  $\tt S=ut+\dfrac{1}{2}at^2$

Substituting the values,

$\longrightarrow \rm 0.01=0(t)+\dfrac{1}{2}(\dfrac{9.77}{1}) t^2 \\\\ \longrightarrow \rm 0.01=4.885 \times t^2 \\\\ \longrightarrow \rm t^2=\dfrac{0.01}{4.885} \\\\ \longrightarrow \rm t^2 \approx 0.205 \times 10^{-2} \\\\ \longrightarrow \rm t=0.453 \times 10^{-1} \: s \\\\ \longrightarrow \rm t=0.0453 \: s$

∴ The time taken by A to fall through 1 cm is 0.0453 s.

Answer 2

Given:

mA = 1 kg, mB = 100 kg, d = 100 m.

(i) It's known to us that, F = GMm/d^2

So, F = 6.67 × 10^-11 × (100)/(10000) N

=> F = 6.67 × 10^-11-2 = 6.67 × 10^-13 N

(ii) Assume that the sphere (A) is kept on the surface of earth. The force by earth to the sphere is called Weight. It is given by, W = mg.

Therefore,

W = mA × g = 1 kg × 9.8 m/s^2 = 9.8 N

(iii) Assume they are kept in an idle system.

Here, g (acceleration due to B's gravity) = 6.67 × 10^-13 m/s² (From W = mg where m = 1)

u = 0 m/s, t = 1

Since, v = u + gt,

v = g(1 s) = 6.67 × 10^-13 m/s

Velocity of A increases with time.

(iv) u = 0 m/s, g = 9.8 m/s^2, t = 1.

So, v = gt = 9.8 m/s.

Now, S = 1/100 m, t = ?, u = 0 m/s, g = 9.8 m/s^2

Since, S = ut + 1/2 gt^2,

S = 1/2 gt^2

=> 1/100 = 1/2 × 9.8 t^2

=> 4.9 t^2 = 1/100

=> 49 t^2 = 1/10

=> t = 1/7√10 or 0.045 s approximately.

More:

• g at height h:

g′ = g(R/R+h)²

• g at depth d:

g′ = g(1 - d/R).