At what height above the Earth’s surface would the Earth’s gravitational field strength be equal to 7.5 N/kg?

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At what height above the Earth’s surface would the Earth’s gravitational field strength be equal to 7.5 N/kg?

Answer given: 915.8 km

2 Answers

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Answer 1 · 2016-12-18T19:53:44

This occurs at a distance of $7.3 \times 10^{6}$ $7.3xx10^6$ m from the centre of the Earth (which is 900 km from the surface).

Explanation:

Are we okay to start from the field equation for the Earth's gravitational field?

$g = \frac{G M}{r^{2}}$ $g=(GM)/r^2$ where M is the mass of the Earth.

What you do now, is change the subject of this equation so it solves for r:

$r = \sqrt{\frac{G M}{g}}$ $r = sqrt((GM)/g)$

Plug in the numbers, and you have it!

$r = \sqrt{\frac{(6.67 \times 10^{- 11}) (5.98 \times 10^{24})}{7.5}}$ $r = sqrt(((6.67xx10^(-11))(5.98xx10^(24)))/7.5)$

= $7.3 \times 10^{6}$ $7.3 xx 10^6$ m

(By the way, since this value is from the centre of the Earth, and the radius of the Earth is $6.4 \times 10^{6}$ $6.4 xx 10^6$ m, we are looking at a point only 900 000 m (900 km) above the surface of the Earth.)

Answer 2 · 2016-12-18T19:56:56

See below.

Explanation:

Given

$G = 6.673 10^{- 11}$ $G = 6.673 10^(-11)$
$M_{e a r t h} = 5.98 10^{24}$ $M_(earth) = 5.98 10^24$
$R_{e a r t h} = 6.38 10^{6}$ $R_(earth) = 6.38 10^6$ All in SI units.

We need the value of $r$ $r$ such that

$\frac{G M_{e a r t h}}{{(R_{e a r t h} + r)}_{e a r t h}^{2}} = 7.5$ $(G M_(earth))/(R_(earth)+r)^2=7.5$

Solving for $r$ $r$ we get

$r = 914.2$ $r = 914.2$ Km

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At what height above the Earth’s surface would the Earth’s gravitational field strength be equal to 7.5 N/kg?

Answer given: 915.8 km

2 Answers

Explanation:

Explanation:

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