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The position vector of the objects of ma...

The position vector of the objects of masses 25 kg and 10 kg are (4,7,5)m and (1,3,5) m respectively. Obtain the vector representing the gravitational force on 25 kg object by 10 kg object. (Take `G = 6.67 xx 10^(-11) Nm^(2)kg^(-2) )`

Text Solution

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`implies` Here , `m_1 = 25 kg , m_2 = 10 kg `
`vecr_1 = (4,7,5) m, vecr_2 = (1,3,5) m, vecF_(12) = (?)`
`vecF_(12) = G(m_1m_2)/(r^2) hatr_(12) " "....(1)`
`vecr_(12) = vecr_(2) - vecr_(1)= (1,3,5) - (4,7,5) = (-3, -4,0) m `
`:. r = |vecr_(12)|=sqrt((-3)^2+(-4)^2+(0)^2)=5m`
and ` vecr_(12) = (vecr_(12))/(|vecr_(12)|)=((-3,-4,0))/5`
`= (0.6, -0.8 ,0)`
`= 0.6 hati - 0.8 hatj`
Putting this in equation (1) ,
`vecF_(12)=(6.67 xx10^(-11))((25)(10))/(5^2) xx(-0.6 , 0.8 , 0)`
`= 6.67 xx 10^(-10)(0.6 hati - 0.8 hatj) N`
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Knowledge Check

  • A particle of mass 10 g is kept of the surface of a uniform sphere of mass 100 kg and radius 10 cm. Find the work to be done against the gravitational force between them to take the particle far away from the sphere. (Take G = 6.67 xx 10^(-11) Nm^(2)//kg^(2) ?

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    D
    `3.33xx 10^(-10) J`
  • The ratio of time taken by two objects A and B of masses 1 kg and 3 kg to free fall from height 16 m and 25 m respectively is ......

    A
    `12/5`
    B
    `5/12`
    C
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    D
    `5/4`
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