Question

the center of the hubble space telescope is 6940 km from earths center. if the gravitational force between earth and the telescope is $9.21 \times 10^4$ n, and the mass of earth is $5.98 \times 10^{24}$ kg, what is the mass of the telescope? round the answer to the nearest whole number.
\boxed{1} kg

Explanation:

Step1: Recall the gravitational force formula

The gravitational force formula is \( F = G\frac{m_1m_2}{r^2} \), where \( F \) is the gravitational force, \( G \) is the gravitational constant (\( G = 6.674\times 10^{-11}\ N\cdot m^2/kg^2 \)), \( m_1 \) and \( m_2 \) are the masses of the two objects, and \( r \) is the distance between their centers. We need to solve for \( m_2 \) (mass of the telescope), so we can rearrange the formula to \( m_2=\frac{F r^2}{G m_1} \).

Step2: Convert the distance to meters

The distance \( r = 6940\ km \). Since \( 1\ km = 1000\ m \), we have \( r = 6940\times1000 = 6.94\times 10^{6}\ m \).

Step3: Substitute the values into the formula

We know that \( F = 9.21\times 10^{4}\ N \), \( r = 6.94\times 10^{6}\ m \), \( G = 6.674\times 10^{-11}\ N\cdot m^2/kg^2 \), and \( m_1 = 5.98\times 10^{24}\ kg \).

Substitute these values into the formula for \( m_2 \):

\[ \begin{align*} m_2&=\frac{(9.21\times 10^{4})\times(6.94\times 10^{6})^2}{(6.674\times 10^{-11})\times(5.98\times 10^{24})}\\ \end{align*} \]

First, calculate the numerator: \( (9.21\times 10^{4})\times(6.94\times 10^{6})^2=(9.21\times 10^{4})\times(4.81636\times 10^{13}) = 9.21\times4.81636\times 10^{4 + 13}=44.36\times 10^{17}=4.436\times 10^{18} \) (approximate calculation for intermediate step, more precise calculation below)

More precisely, \( (6.94\times 10^{6})^2=6.94^{2}\times 10^{12}=48.1636\times 10^{12}=4.81636\times 10^{13} \)

Then \( 9.21\times 10^{4}\times4.81636\times 10^{13}=9.21\times4.81636\times 10^{4 + 13}=44.36\times 10^{17}=4.436\times 10^{18} \) (exact: \( 9.21\times4.81636 = 9.21\times4 + 9.21\times0.81636=36.84+7.52=44.36 \) (approx), more accurately \( 9.21\times4.81636 = 44.36 \))

Denominator: \( (6.674\times 10^{-11})\times(5.98\times 10^{24})=6.674\times5.98\times 10^{-11 + 24}=39.91\times 10^{13}=3.991\times 10^{14} \) (more precisely \( 6.674\times5.98 = 39.91 \))

Now, \( m_2=\frac{4.436\times 10^{18}}{3.991\times 10^{14}}=\frac{4.436}{3.991}\times 10^{18 - 14}\approx1.1115\times 10^{4}=11115 \) (approximate). Wait, let's do the calculation more accurately.

First, calculate numerator:

\( (9.21\times 10^{4})\times(6940\times 1000)^2=(9.21\times 10^{4})\times(6.94\times 10^{6})^2=(9.21\times 10^{4})\times(4.81636\times 10^{13}) = 9.21\times4.81636\times 10^{17} \)

\( 9.21\times4.81636 = 9\times4.81636+0.21\times4.81636 = 43.34724+1.0114356 = 44.3586756 \)

So numerator is \( 44.3586756\times 10^{17}=4.43586756\times 10^{18} \)

Denominator:

\( 6.674\times 10^{-11}\times5.98\times 10^{24}=6.674\times5.98\times 10^{13} \)

\( 6.674\times5.98 = 6\times5.98+0.674\times5.98 = 35.88+4.03052 = 39.91052 \)

So denominator is \( 39.91052\times 10^{13}=3.991052\times 10^{14} \)

Now, \( m_2=\frac{4.43586756\times 10^{18}}{3.991052\times 10^{14}}=\frac{4.43586756}{3.991052}\times 10^{4} \)

\( \frac{4.43586756}{3.991052}\approx1.1115 \)

So \( 1.1115\times 10^{4}=11115 \) (when rounded to the nearest whole number)

Wait, let's check the calculation again. Maybe I made a mistake in exponent.

Wait, \( r = 6940\ km = 6940\times 1000\ m=6.94\times 10^{6}\ m \), so \( r^2=(6.94\times 10^{6})^2 = 6.94^2\times 10^{12}=48.1636\times 10^{12}=4.81636\times 10^{13}\ m^2 \)

\( F = 9.21\times 10^{4}\ N \)

\( G = 6.674\times 10^{-11}\ Nm^2/kg^2 \)

\( m_1 = 5.98\times 10^{24}\ kg \)

So \( m_2=\frac{F r^2}{G m_1}=\frac{9.21\times 10^{4}\times4.81636\times 10^{13}}{6.674\times 10^{-11}\times5.98\times 10^{24}} \)

Combine the exponents in numerator: \( 10^{4 + 13}=10^{17} \)

In denominator: \( 10^{-11+24}=10^{13} \)

So now the formula becomes \( \frac{9.21\times4.81636\times 10^{17}}{6.674\times5.98\times 10^{13}}=\frac{9.21\times4.81636}{6.674\times5.98}\times 10^{17 - 13}=\frac{9.21\times4.81636}{6.674\times5.98}\times 10^{4} \)

Calculate the fraction:

\( 9.21\times4.81636 = 44.3586756 \)

\( 6.674\times5.98 = 39.91052 \)

\( \frac{44.3586756}{39.91052}\approx1.1115 \)

Then \( 1.1115\times 10^{4}=11115 \)

Yes, so the mass of the telescope is approximately 11115 kg.

Answer:

11115