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A spacecraft of mass $1.0 \times 10^6$ kg is in orbit around the Sun at a radius of $1.1 \times 10^{11}$ m - AQA - A-Level Physics - Question 17 - 2018 - Paper 2

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A-spacecraft-of-mass-$1.0-\times-10^6$-kg-is-in-orbit-around-the-Sun-at-a-radius-of-$1.1-\times-10^{11}$-m-AQA-A-Level Physics-Question 17-2018-Paper 2.png

A spacecraft of mass $1.0 \times 10^6$ kg is in orbit around the Sun at a radius of $1.1 \times 10^{11}$ m. The spacecraft moves into a new orbit of radius $2.5 \tim... show full transcript

Worked Solution & Example Answer:A spacecraft of mass $1.0 \times 10^6$ kg is in orbit around the Sun at a radius of $1.1 \times 10^{11}$ m - AQA - A-Level Physics - Question 17 - 2018 - Paper 2

Step 1

Calculate the initial gravitational potential energy

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Answer

The gravitational potential energy (U) is given by the formula:

U=GMmrU = -\frac{G M m}{r}

Where:

  • G is the gravitational constant (approximately 6.674×1011 m3 kg1 s26.674 \times 10^{-11} \ m^3 \ kg^{-1} \ s^{-2})
  • M is the mass of the Sun (approximately 1.989×1030 kg1.989 \times 10^{30} \ kg)
  • m is the mass of the spacecraft (1.0×106 kg1.0 \times 10^6 \ kg)
  • r is the distance from the center of the Sun to the spacecraft.

For the initial radius (r1=1.1×1011 mr_1 = 1.1 \times 10^{11} \ m):

U1=(6.674×1011)(1.989×1030)(1.0×106)1.1×1011 U_1 = -\frac{(6.674 \times 10^{-11}) (1.989 \times 10^{30}) (1.0 \times 10^6)}{1.1 \times 10^{11}}

Step 2

Calculate the final gravitational potential energy

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Answer

For the final radius (r2=2.5×1011 mr_2 = 2.5 \times 10^{11} \ m):

U2=(6.674×1011)(1.989×1030)(1.0×106)2.5×1011 U_2 = -\frac{(6.674 \times 10^{-11}) (1.989 \times 10^{30}) (1.0 \times 10^6)}{2.5 \times 10^{11}}

Step 3

Determine the change in gravitational potential energy

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Answer

The change in gravitational potential energy (ΔU\Delta U) is calculated as:

ΔU=U2U1\Delta U = U_2 - U_1

Substituting the values calculated from the previous steps will yield the total change.

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