3 years ago

. The total energy when a body falls freely towards earth remains constant. True or false?

Physics

1 answer:

Colt1911 [192]3 years ago

7 0

This is true, The total energy when a body falls freely towards earth remains constant.

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3 years ago

The drawing (not to scale) shows one alignment of the sun, earth, and moon. the gravitational force vector f sm that the sun exe

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Solution:

Ms = 1.99 × 1030 kg− mass of Sun;

Me = 5.98 × 1024kg− mass of Earth;

Mm = 7.35 × 1022kg − mass of Moon;

rSM = 1.50 × 1011m − distance to the Moon from the Sun;

rEM = 3.85 × 108m − distance to the Moon from the Earth;

The gravitational force that acts on the Moon by the Earth (Law of Gravity):

$F_{e} = G \frac {M_{e} * M_{m} } {r^{2}_{EM}} = 6.67 x
10^{-11} N * (\frac {m} {kg})^{2}*\frac {5.98 * 10^{24} kg * 7.35 * 10^{22} kg}
{(3.85 x 10^{8}m)^{2}} = 1.98 * 10^{20} N$

The gravitational force that acts on the Moon by the Sun (Law of Gravity):

$F_{S} = G \frac {M_{s} * M_{m} } {r^{2}_{EM}} = 6.67 x
10^{-11} N * (\frac {m} {kg})^{2}*\frac {1.99 * 10^{30} kg * 7.35 * 10^{22} kg}
{(1.50 x 10^{11}m)^{2}} = 4.34 * 10^{20} N$

Net gravitational force on the moon:

$F = F_{e} + F_{s}$

Pythagorean theorem for a right triangle ABC:

$F = \sqrt {F^{2}_{S} + F^{2}_{e}} = \sqrt {(1.98 *
10^{20}N)^{2} + (4.34 * 10^{20} N)^{2}} = 4.77 * 10^{20}N$

Answer: Answer: magnitude of the net gravitational force on the moon is 4.77 × $10^{20}$ N.