Question

1. what is the gravitational force between an object of mass $3.6 \times 10^2$ kg and a second object of mass $4.3 \times 10^3$ kg when the distance between their centres is 53 m? (6.1) t/i

(a) $3.7 \times 10^{-8}$ n
(b) $1.9 \times 10^{-6}$ n
(c) $5.4 \times 10^{-3}$ n
(d) $2.9 \times 10^{-1}$ n

Explanation:

Step1: Recall Gravitational Force Formula

The formula for gravitational force between two objects is \( F = G\frac{m_1m_2}{r^2} \), where \( G = 6.67\times 10^{-11}\ Nm^2/kg^2 \), \( m_1 \) and \( m_2 \) are the masses of the two objects, and \( r \) is the distance between their centers.

Step2: Identify Given Values

Here, \( m_1 = 3.6\times 10^{2}\ kg \), \( m_2 = 4.3\times 10^{3}\ kg \), and \( r = 53\ m \).

Step3: Substitute Values into Formula

First, calculate the product of the masses: \( m_1m_2=(3.6\times 10^{2})\times(4.3\times 10^{3}) = 3.6\times4.3\times 10^{2 + 3}=15.48\times 10^{5}=1.548\times 10^{6}\ kg^2 \).
Then, calculate \( r^2=(53)^2 = 2809\ m^2 \).
Now, substitute into the formula: \( F=(6.67\times 10^{-11})\times\frac{1.548\times 10^{6}}{2809} \).
First, calculate \( \frac{1.548\times 10^{6}}{2809}\approx\frac{1.548\times 10^{6}}{2.809\times 10^{3}}\approx0.551\times 10^{3}=5.51\times 10^{2} \).
Then, \( F = 6.67\times 10^{-11}\times5.51\times 10^{2}=6.67\times5.51\times 10^{-11 + 2} \).
Calculate \( 6.67\times5.51\approx36.75 \), so \( F\approx36.75\times 10^{-9}=3.675\times 10^{-8}\ N\approx3.7\times 10^{-8}\ N \).

Answer:

(a) \( 3.7\times 10^{-8}\ N \)